Real Numbers & Exponents

Almost everyone who struggles in algebra is not actually struggling with algebra — they are struggling with a handful of arithmetic rules that were taught quickly, years ago, and never revisited. This chapter goes back and sets those rules down carefully: what kinds of numbers exist, how they behave when you add and multiply them, what order operations happen in, and how exponents work. Nothing here is beyond you, and nothing here will be rushed. Every rule gets a reason, and every example gets worked out one visible step at a time. Get this chapter solid and the rest of algebra stops feeling like guesswork.

Section 1The real number system

Start with counting, then notice what is missing

The first numbers anyone learns are the ones you count with: 1, 2, 3, 4, and onward forever. These are the natural numbers (often written N), sometimes called the counting numbers. They are enormously useful and badly incomplete. Try to count the sheep in an empty field and you need a number the naturals do not contain. Add zero and you have the whole numbers (W): 0, 1, 2, 3, and up.

Now try to record a temperature of eight degrees below zero, or a checking account that is 40 dollars overdrawn. You need numbers on the other side of zero. Attach the negatives and you have the integers (Z): …, −3, −2, −1, 0, 1, 2, 3, … Notice the pattern already forming. Each new set keeps everything the old set had and adds the piece that was missing. Every natural number is a whole number; every whole number is an integer. That nesting is exactly what the figure above is showing you.

Rational numbers: anything you can write as a fraction

Integers still cannot handle half a pizza. A rational number (Q, for quotient) is any number that can be written as a ratio of two integers — that is, as (a)/(b) where a and b are integers and b ≠ 0. That restriction is not decoration. Division by zero is undefined, so a denominator of zero is never allowed, and you will see that condition attached to fractions for the rest of your mathematical life.

Rationals are a bigger group than they first appear. Every integer is secretly a fraction: 7 = (7)/(1), and −12 = (−12)/(1). Every terminating decimal is a fraction: 0.25 = (25)/(100) = (1)/(4). And — this one surprises people — every repeating decimal is a fraction too. In decimal form, a rational number either stops or falls into a repeating block. There is no third option.

Irrational numbers, and the real number line

Some decimals never stop and never repeat. Those are the irrational numbers. The famous ones are √2 = 1.41421356…, π = 3.14159265…, and e = 2.71828182…, and none of them can be written as a fraction of two integers — that is a fact that can be proved, not just a failure to find the right fraction. Careful: √9 = 3 is perfectly rational. It is only the square roots that do not come out evenly that land in the irrational pile.

Put the rationals and the irrationals together and you get the real numbers (R) — every number that has a place on the number line, with no gaps anywhere. Every real number is exactly one point on that line, and every point on that line is exactly one real number. One last restriction worth knowing now: √−4 is not a real number, because no real number multiplied by itself gives a negative result. Even roots of negatives step outside the real number system entirely.

Worked example: turning a repeating decimal into a fraction

Here is the trick that proves repeating decimals are rational. Let us convert 0.727272… step by step, showing every line.

Step 1 — name the number. Let x = 0.727272…

Step 2 — shift the decimal past one full repeating block. The block "72" is two digits long, so multiply both sides by 10² = 100:
100x = 72.727272…

Step 3 — subtract the original equation from the new one. The infinite tails are identical, so they cancel:
100x − x = 72.727272… − 0.727272…
99x = 72

Step 4 — solve for x. Divide both sides by 99:
x = (72)/(99)

Step 5 — reduce. Both 72 and 99 are divisible by 9. 72 ÷ 9 = 8 and 99 ÷ 9 = 11, so
x = (8)/(11)

Step 6 — check. Divide 8 by 11 by hand: 8.000 ÷ 11 gives 0.7272… ✓ The original decimal comes back, so the answer is right. And since 8 and 11 are both integers with 11 ≠ 0, the number 0.727272… is rational — exactly as promised.

Section 2Properties of real numbers

Five rules you already use without naming

You have been using these properties for years without ever calling them by name. The reason to learn the names now is that textbooks, instructors, and test questions will ask you to justify a step, and "because that's how it works" is not an answer they accept. Each property is a promise about what you are allowed to rearrange without changing the value.

The commutative property says order does not matter: a + b = b + a, and a · b = b · a. The associative property says grouping does not matter: (a + b) + c = a + (b + c), and (ab)c = a(bc). The distributive property is the one that connects multiplication to addition: a(b + c) = ab + ac. It is the single most-used property in all of algebra, and it runs in both directions — expanding when you read it left to right, factoring when you read it right to left.

The identity properties name the two numbers that change nothing: a + 0 = a (zero is the additive identity) and a · 1 = a (one is the multiplicative identity). The inverse properties name the partner that undoes a number: a + (−a) = 0 (the opposite, or additive inverse) and a · (1)/(a) = 1 (the reciprocal, or multiplicative inverse) — and that second one carries a restriction, a ≠ 0, because (1)/(0) does not exist.

Where these properties do not apply

This is where careless work goes wrong. Commutative and associative apply to addition and multiplication only. Subtraction and division get neither. Check it yourself with real numbers: 8 − 3 = 5 but 3 − 8 = −5, so subtraction is not commutative. And 12 ÷ 4 = 3 but 4 ÷ 12 = (1)/(3), so division is not commutative either. Grouping fails the same way: (12 − 5) − 2 = 7 − 2 = 5, while 12 − (5 − 2) = 12 − 3 = 9. Same numbers, same order, different grouping, different answer.

There is a repair, though, and it is worth internalizing: rewrite subtraction as adding the opposite, and rewrite division as multiplying by the reciprocal. Once 8 − 3 becomes 8 + (−3), you are back in addition and all the addition properties return. This is why algebra teachers keep saying "subtraction is just addition in a costume."

Worked example: naming every property in a simplification

Simplify 3(2x − 5) + 4x, and justify each step by name.

Step 1 — rewrite the subtraction as adding the opposite so the distributive property applies cleanly:
3(2x − 5) + 4x = 3 [ 2x + (−5) ] + 4x

Step 2 — distribute the 3 across both terms (distributive property):
= 3(2x) + 3(−5) + 4x

Step 3 — do the arithmetic inside each product. 3 · 2 = 6, and 3 · (−5) = −15:
= 6x + (−15) + 4x

Step 4 — move 4x next to 6x (commutative property of addition):
= 6x + 4x + (−15)

Step 5 — group the like terms (associative property of addition):
= (6x + 4x) + (−15)

Step 6 — factor out the x (distributive property, read right to left):
= (6 + 4)x + (−15)

Step 7 — finish the arithmetic. 6 + 4 = 10, and "+ (−15)" is written as "− 15":
= 10x − 15

Check with a number. Let x = 2. Original: 3(2·2 − 5) + 4·2 = 3(4 − 5) + 8 = 3(−1) + 8 = −3 + 8 = 5. Simplified: 10(2) − 15 = 20 − 15 = 5. ✓ They match, so the simplification is trustworthy.

The distributive property as a mental-math shortcut

Watch how much work the distributive property saves. To compute 7 × 98 without a calculator, notice that 98 = 100 − 2:

7 × 98 = 7(100 − 2) = 7(100) − 7(2) = 700 − 14 = 686

That is the same property you will use to expand 7(x − 2) into 7x − 14. Numbers and variables obey identical rules — that is really the whole idea behind algebra.

Section 3Order of operations

Why an order is needed at all

Take the expression 2 + 3 × 4. Add first and you get 5 × 4 = 20. Multiply first and you get 2 + 12 = 14. Both are defensible readings, and that is intolerable — an expression has to mean one thing. So mathematicians agreed on a fixed order, and everyone from your calculator to a spacecraft's flight software follows it. The correct value is 14, because multiplication outranks addition.

The agreed order is often remembered as PEMDAS: Parentheses, Exponents, Multiplication and Division, Addition and Subtraction. But that acronym misleads thousands of students every year, so read the next sentence twice. Multiplication and division sit on the same level and are done left to right as they appear; addition and subtraction sit on the same level and are also done left to right. PEMDAS has four levels, not six.

Signed numbers, carefully

Most order-of-operations mistakes are really sign mistakes. Four rules cover nearly everything. When you multiply or divide, same signs give a positive and different signs give a negative: (−6)(−7) = 42, but (−6)(7) = −42. When you add two numbers with the same sign, add their sizes and keep the sign: −5 + (−8) = −13. When you add two numbers with different signs, subtract the smaller size from the larger and keep the sign of the larger: −9 + 4 = −5. And subtracting is adding the opposite: 6 − (−4) = 6 + 4 = 10.

Then there is the single most common exponent-with-signs trap. In −3², the exponent attaches only to the 3, so you square first and negate second: −3² = −(3 · 3) = −9. In (−3)², the parentheses make −3 the base, so (−3)² = (−3)(−3) = 9. Those two expressions differ by more than a sign — they are different questions.

Worked example: a full expression, one step at a time

Evaluate 8 + 3(7 − 2 · 5)² ÷ 9 − (−4). Nothing gets skipped.

Step 1 — work inside the parentheses first. Inside, multiplication outranks subtraction, so 2 · 5 = 10:
= 8 + 3(7 − 10)² ÷ 9 − (−4)

Step 2 — finish inside the parentheses. 7 − 10 = −3:
= 8 + 3(−3)² ÷ 9 − (−4)

Step 3 — exponents. The base is −3 because of the parentheses, and (−3)(−3) = 9:
= 8 + 3 · 9 ÷ 9 − (−4)

Step 4 — multiplication and division, left to right. The leftmost is 3 · 9 = 27:
= 8 + 27 ÷ 9 − (−4)

Step 5 — continue left to right. 27 ÷ 9 = 3:
= 8 + 3 − (−4)

Step 6 — addition and subtraction, left to right. 8 + 3 = 11:
= 11 − (−4)

Step 7 — subtracting a negative is adding a positive. 11 + 4 = 15.

Notice how much of the difficulty lived in Steps 3 and 7 — the two places signs mattered. That is typical.

Worked example: substituting values into an expression

Evaluate b² − 4ac when a = 2, b = −5, and c = −3. (You will meet this exact expression again as the discriminant of the quadratic formula.)

Step 1 — substitute, and put every negative value in parentheses. This one habit prevents most sign errors:
b² − 4ac = (−5)² − 4(2)(−3)

Step 2 — exponents first. (−5)(−5) = 25:
= 25 − 4(2)(−3)

Step 3 — multiply left to right. 4 · 2 = 8:
= 25 − 8(−3)

Step 4 — keep multiplying. 8 · (−3) = −24:
= 25 − (−24)

Step 5 — subtract the negative. 25 + 24 = 49.

Worked example: the fraction bar as a grouping symbol

Evaluate (4 − 2(−3))/((−1)² + 4). A fraction bar silently wraps parentheses around the entire numerator and the entire denominator, so both must be finished before you divide.

Step 1 — numerator, multiplication first. 2(−3) = −6, so the numerator is 4 − (−6) = 4 + 6 = 10.

Step 2 — denominator, exponent first. (−1)² = (−1)(−1) = 1, so the denominator is 1 + 4 = 5.

Step 3 — now divide. (10)/(5) = 2.

Before dividing, always confirm the denominator is not zero. Here it is 5, so the expression is defined. If a denominator ever works out to 0, the expression is undefined and there is no value to report.

Section 4Integer exponents

What an exponent actually abbreviates

An exponent is nothing but repeated multiplication written compactly. In a⁵, the number a is the base and 5 is the exponent or power, and it means a · a · a · a · a — five factors of a. Every exponent rule you are about to learn can be rediscovered by writing out the factors and cancelling. That is worth knowing, because it means you never have to memorize a rule you cannot reconstruct.

Watch the product rule appear on its own. Take a³ · a² and expand: (a · a · a)(a · a) = a · a · a · a · a = a⁵. Three factors plus two factors is five factors, so a^m · aⁿ = a^m+n. You add exponents when multiplying like bases — and note "like bases" is required. a³ · b² cannot be combined at all.

The zero and negative exponent rules are not arbitrary

Students are often told "anything to the zero power is 1" with no explanation, and it feels like a made-up rule. It is not. Compute (2³)/(2³) two different ways. By plain arithmetic, (8)/(8) = 1. By the quotient rule, 2³⁻³ = 2⁰. Both answers describe the same quantity, so 2⁰ = 1. The same argument works for any nonzero base, which is exactly why the rule carries the restriction a ≠ 0 — 0⁰ is undefined, because the argument would require dividing 0 by 0.

Negative exponents come from the same move. Compute (2³)/(2⁵) both ways. Arithmetic: (8)/(32) = (1)/(4). Quotient rule: 2³⁻⁵ = 2⁻². So 2⁻² = (1)/(4) = (1)/(2²). In general a⁻ⁿ = (1)/(aⁿ) with a ≠ 0. Say this out loud once and it will stick: a negative exponent does not make a number negative — it moves the factor to the other side of the fraction bar. So 2⁻³ = (1)/(8), a positive number, and (1)/(x⁻⁴) = x⁴.

Worked example: simplifying with negative exponents

Simplify (12x⁵y⁻³)/(3x⁻²y⁴) and write the answer with positive exponents only. State any restrictions.

Step 0 — state restrictions before you touch anything. The variables sit in denominators (and y⁻³ is itself a denominator in disguise), so we need x ≠ 0 and y ≠ 0. Everything below assumes that.

Step 1 — separate the expression into number, x, and y pieces:
= (12)/(3) · (x⁵)/(x⁻²) · (y⁻³)/(y⁴)

Step 2 — divide the numbers. 12 ÷ 3 = 4:
= 4 · (x⁵)/(x⁻²) · (y⁻³)/(y⁴)

Step 3 — apply the quotient rule to x. Subtract the exponents, and be careful with the double negative: 5 − (−2) = 5 + 2 = 7:
= 4x⁷ · (y⁻³)/(y⁴)

Step 4 — apply the quotient rule to y. −3 − 4 = −7:
= 4x⁷y⁻⁷

Step 5 — clear the negative exponent by moving that factor to the denominator:
= (4x⁷)/(y⁷)    for x ≠ 0, y ≠ 0

Check with numbers. Let x = 1 and y = 2. Original: (12 · 1 · 2⁻³)/(3 · 1 · 2⁴) = (12 · (1)/(8))/(3 · 16) = (1.5)/(48) = 0.03125. Answer: (4 · 1)/(2⁷) = (4)/(128) = 0.03125. ✓

Worked example: a negative power of a whole group

Simplify (2a³b⁻¹)⁻², positive exponents only, with a ≠ 0 and b ≠ 0.

Step 1 — the outer exponent hits every factor inside (power of a product):
= 2⁻² · (a³)⁻² · (b⁻¹)⁻²

Step 2 — multiply the exponents (power of a power). 3 · (−2) = −6 and (−1)(−2) = 2:
= 2⁻² · a⁻⁶ · b²

Step 3 — evaluate 2⁻². It equals (1)/(2²) = (1)/(4):
= (1)/(4) · a⁻⁶ · b²

Step 4 — move a⁻⁶ to the denominator and collect:
= (b²)/(4a⁶)    for a ≠ 0, b ≠ 0

A quick sanity check on Step 3: the answer must be positive, since 2⁻² is a positive fraction. If you had written −4 there, the negative exponent trap caught you.

Section 5Scientific notation

The form, and the one rule that defines it

Real quantities are frequently absurd to write out. The mass of Earth is about 5,970,000,000,000,000,000,000,000 kilograms; the diameter of a hydrogen atom is about 0.000000000106 meters. Counting those zeros is an invitation to error, so we use scientific notation: every number is written as

a × 10ⁿ,   where   1 ≤ |a| < 10   and   n is an integer

That condition on a is the whole rule, and it is strict: exactly one nonzero digit sits to the left of the decimal point. So 5.97 × 10²⁴ is proper scientific notation, while 59.7 × 10²³ and 0.597 × 10²⁵ are equal in value but not in proper form. The exponent n is positive for large numbers (|a × 10ⁿ| of 10 or more), zero when the number itself already sits between 1 and 10, and negative for small ones (absolute value between 0 and 1), which gives you an instant sanity check on any answer.

Worked example: converting in both directions

Large number. Write 5,970,000,000,000,000,000,000,000 in scientific notation.

Step 1 — place the decimal so exactly one nonzero digit is in front of it. That puts it between the 5 and the 9: 5.97

Step 2 — count how many places the decimal moved. It started at the far right end of the number and moved left 24 places.

Step 3 — moving left means a positive exponent (you shrank a huge number, so the power of ten must build it back up):
5,970,000,000,000,000,000,000,000 = 5.97 × 10²⁴ kg

Small number. Write 0.000000000106 in scientific notation.

Step 1 — place the decimal after the first nonzero digit. The first nonzero digit is 1, so the number becomes 1.06

Step 2 — count the move. From 0.000000000106 the decimal moved right 10 places to sit after the 1.

Step 3 — moving right means a negative exponent:
0.000000000106 = 1.06 × 10⁻¹⁰ m

Going back the other way. To expand 3.4 × 10⁻⁵, move the decimal 5 places left, filling with zeros: 0.000034. To expand 2.9 × 10⁶, move 6 places right: 2,900,000.

Worked example: multiplying and dividing

This is where Section 4 pays off — the exponent rules do all the work.

Multiply: (3.2 × 10⁵)(4.5 × 10⁻⁹)

Step 1 — regroup so the decimals are together and the powers of ten are together (commutative and associative properties, doing real work):
= (3.2 × 4.5)(10⁵ × 10⁻⁹)

Step 2 — multiply the decimals. 3.2 × 4.5 = 14.4

Step 3 — add the exponents (product rule). 5 + (−9) = −4:
= 14.4 × 10⁻⁴

Step 4 — fix the form. 14.4 is not between 1 and 10, so rewrite 14.4 as 1.44 × 10¹:
= 1.44 × 10¹ × 10⁻⁴ = 1.44 × 10⁻³

Divide: (6.4 × 10⁷)/(8.0 × 10⁻³)

Step 1 — split into a decimal quotient and a power-of-ten quotient:
= (6.4)/(8.0) × (10⁷)/(10⁻³)

Step 2 — divide the decimals. 6.4 ÷ 8.0 = 0.8

Step 3 — subtract the exponents (quotient rule), watching the double negative: 7 − (−3) = 7 + 3 = 10:
= 0.8 × 10¹⁰

Step 4 — fix the form. 0.8 is less than 1, so write it as 8.0 × 10⁻¹:
= 8.0 × 10⁻¹ × 10¹⁰ = 8.0 × 10⁹

Worked example: adding requires a matching exponent

Add 4.2 × 10⁶ + 3.5 × 10⁵. You cannot simply add the decimals here, because the two terms are not measured in the same size unit — one is in millions, the other in hundred-thousands.

Step 1 — rewrite the smaller term to match the larger exponent. To raise the exponent from 5 to 6, shift the decimal one place left: 3.5 × 10⁵ = 0.35 × 10⁶. (Check: 0.35 × 1,000,000 = 350,000, and 3.5 × 100,000 = 350,000. ✓)

Step 2 — now both terms carry 10⁶, so factor it out (distributive property):
= (4.2 + 0.35) × 10⁶

Step 3 — add the decimals. 4.2 + 0.35 = 4.55:
= 4.55 × 10⁶

Step 4 — confirm the form. 4.55 sits between 1 and 10, so the answer is already proper. Written out, that is 4,550,000 — which matches 4,200,000 + 350,000. ✓

ReferenceWords worth knowing

A quick reference for the vocabulary in this chapter — skim it now, and come back whenever a word in a problem statement slows you down.

Absolute value

The distance a number sits from zero on the number line, always zero or positive: |−7| = 7 and |7| = 7. The bars also act as a grouping symbol.

Associative property

Regrouping does not change the result: (a+b)+c = a+(b+c) and (ab)c = a(bc). Applies to addition and multiplication only.

Base

In aⁿ, the number a being repeatedly multiplied. In −3² the base is 3; in (−3)² the base is −3.

Commutative property

Reordering does not change the result: a+b = b+a and ab = ba. Applies to addition and multiplication only.

Distributive property

a(b+c) = ab+ac. Read left to right it expands; read right to left it factors.

Exponent

The small raised number telling how many factors of the base to multiply; also called the power.

Identity property

Adding 0 or multiplying by 1 leaves a number unchanged: a+0 = a, a·1 = a.

Integers

The whole numbers together with their negatives: …, −2, −1, 0, 1, 2, …

Irrational number

A real number whose decimal never terminates and never repeats, so it cannot be written as a ratio of integers — for example π, e, and √2.

Multiplicative inverse (reciprocal)

The number that multiplies with a to give 1, namely (1)/(a). Defined only for a ≠ 0.

Natural numbers

The counting numbers 1, 2, 3, 4, … — no zero, no negatives, no fractions.

Opposite (additive inverse)

The number that adds with a to give 0, namely −a. Every real number has one, including 0 itself.

Order of operations

The agreed sequence for evaluating an expression: grouping, then exponents, then multiplication and division left to right, then addition and subtraction left to right.

Power

Either the exponent itself or the whole expression aⁿ, depending on context; "3 to the fourth power" means 3⁴ = 81.

Radical sign

The symbol √ indicating a root. It acts as a grouping symbol, so everything under it is evaluated before the root is taken.

Rational number

Any number expressible as (a)/(b) with a and b integers and b ≠ 0; equivalently, any terminating or repeating decimal.

Real number

Any number with a location on the number line — every rational and every irrational number together.

Repeating decimal

A decimal in which a block of digits recurs forever, such as 0.727272… Every repeating decimal is rational.

Scientific notation

Writing a number as a × 10ⁿ with 1 ≤ |a| < 10 and n an integer.

Undefined

Having no value in the given number system — as with division by zero, 0⁰, or √−4 within the real numbers.

Whole numbers

The natural numbers together with 0: 0, 1, 2, 3, …

Linear Equations & Inequalities

Almost everything you will ever do in algebra rests on one small, forgiving idea: whatever you do to one side of an equation, do the same thing to the other side, and the equation stays true. That single rule — the balance rule — is enough to solve every linear equation there is, from x + 3 = 7 to a formula bristling with letters. This chapter walks you through it slowly and out loud: how to isolate a variable, what to do when fractions show up, what it means when an equation has no answer or every answer, how to rearrange a formula for the letter you actually want, how to turn an English sentence into an equation, and how inequalities behave almost identically — with one famous exception. If algebra has ever felt like a bag of tricks, this is the chapter where it becomes a single method you can trust.

Section 1Solving linear equations

What "solve" actually means

An equation is a claim that two expressions are equal. A linear equation in one variable is one where the variable appears only to the first power — no x², no √x, no x in a denominator by itself. To solve it is to find every number you could put in place of the variable that makes the claim true. That number is called a solution (or a root), and the collection of all of them is the solution set.

Here is the part that trips people up, so let us say it plainly: you are not "moving things around." You are performing the same operation on both sides of a balance. If the two sides start out equal and you add 5 to each of them, they are still equal. If you divide each of them by 3, they are still equal. Each such move produces a new equation with exactly the same solutions — an equivalent equation — and you keep making moves until the equation reads x = something.

Notice the word nonzero. Multiplying both sides by zero turns any equation into 0 = 0, which is true but useless; dividing by zero is not defined at all. Those are the only two forbidden moves in this whole chapter.

A worked example, every single step

Solve 5(x − 3) + 2 = 3x + 7. We will not skip anything.

1. Clear the parentheses using the distributive property. 5(x − 3) means 5·x − 5·3 = 5x − 15, so the equation becomes 5x − 15 + 2 = 3x + 7.
2. Combine like terms on each side separately. On the left, −15 + 2 = −13, so we have 5x − 13 = 3x + 7.
3. Collect the variable terms on one side. Subtract 3x from both sides: 5x − 3x − 13 = 3x − 3x + 7, which simplifies to 2x − 13 = 7.
4. Collect the constants on the other side. Add 13 to both sides: 2x − 13 + 13 = 7 + 13, which gives 2x = 20.
5. Undo the coefficient. Divide both sides by 2: (2x)/(2) = (20)/(2), so x = 10.
6. Check. Put 10 back into the original equation. Left side: 5(10 − 3) + 2 = 5(7) + 2 = 35 + 2 = 37. Right side: 3(10) + 7 = 30 + 7 = 37. Both sides are 37, so x = 10 is correct.

That order — parentheses, combine, variables one way, numbers the other way, divide, check — works every time. It is worth memorizing as a rhythm rather than a rule.

Fractions, decimals, and denominators that hold variables

Fractions are not harder, they are just noisier. The fix is to multiply every term on both sides by the least common denominator (LCD), which wipes the denominators out in one stroke. Solve (x)/(3) + (1)/(4) = (x)/(2) − (5)/(6).

1. Find the LCD of 3, 4, 2, and 6. The smallest number all four divide into is 12.
2. Multiply every term by 12. Term by term: 12·(x)/(3) = 4x; 12·(1)/(4) = 3; 12·(x)/(2) = 6x; 12·(5)/(6) = 10. The equation is now 4x + 3 = 6x − 10 — no fractions left.
3. Collect variables. Subtract 4x from both sides: 3 = 2x − 10.
4. Collect constants. Add 10 to both sides: 13 = 2x.
5. Divide by 2. x = (13)/(2) = 6.5.
6. Check with x = 6.5. Left: (6.5)/(3) + (1)/(4) = 2.1667 + 0.25 = 2.4167. Right: (6.5)/(2) − (5)/(6) = 3.25 − 0.8333 = 2.4167. They match, so x = (13)/(2) is correct.

Decimals work the same way in reverse: multiply both sides by 10, 100, or 1000 — whatever clears the longest decimal — and you are back to whole numbers.

If the variable ever sits under a fraction bar, you must first write down the values that would make a denominator zero. Those are restrictions (also called excluded values), and they are off the table before you solve a single step. Solve (3)/(x − 2) = (5)/(x + 4).

1. Restrictions. x − 2 = 0 when x = 2, and x + 4 = 0 when x = −4. So x ≠ 2 and x ≠ −4.
2. Multiply both sides by (x − 2)(x + 4), the LCD. The left becomes 3(x + 4) and the right becomes 5(x − 2), so 3(x + 4) = 5(x − 2).
3. Distribute. 3x + 12 = 5x − 10.
4. Collect. Subtract 3x: 12 = 2x − 10. Add 10: 22 = 2x. Divide by 2: x = 11.
5. Check against the restrictions and the original. 11 is neither 2 nor −4, so it is allowed. Left: (3)/(11 − 2) = (3)/(9) = (1)/(3). Right: (5)/(11 + 4) = (5)/(15) = (1)/(3). Correct.

Sometimes the algebra hands you a number that a restriction forbids. Solve (x)/(x − 3) = (3)/(x − 3) + 4, where x ≠ 3. Multiply everything by (x − 3): x = 3 + 4(x − 3) = 3 + 4x − 12 = 4x − 9. Subtract 4x: −3x = −9, so x = 3. But x = 3 is exactly the excluded value — it makes the original denominators zero. It is an extraneous solution, and the honest answer is that this equation has no solution.

Section 2Identities and contradictions

Not every equation has exactly one answer

Most of the equations you meet have exactly one solution, and it is easy to assume they all do. They do not. A linear equation in one variable has exactly one of three possible outcomes, and which one you get announces itself in the last line of your work.

A conditional equation is true only on the condition that the variable takes one particular value. That is the familiar case: you end with x = 7, and the solution set is {7}. An identity is true no matter what the variable is; the two sides were secretly the same expression all along. You end with something like 4 = 4, and the solution set is all real numbers. A contradiction is true for no value at all; you end with something plainly false like −15 = 1, and the solution set is empty, written ∅ or { }.

Three worked examples, side by side

(a) An identity. Solve 4(x + 1) − 2x = 2x + 4.

1. Distribute on the left: 4x + 4 − 2x = 2x + 4.
2. Combine like terms on the left: 4x − 2x = 2x, so 2x + 4 = 2x + 4.
3. Subtract 2x from both sides: 4 = 4.
4. The variable vanished and what is left is true. This is an identity; every real number works. Spot-check with x = 5: left = 4(6) − 10 = 24 − 10 = 14; right = 10 + 4 = 14. Now try x = −1: left = 4(0) + 2 = 2; right = −2 + 4 = 2. Both work, as promised.

(b) A contradiction. Solve 3(2x − 5) = 6x + 1.

1. Distribute on the left: 6x − 15 = 6x + 1.
2. Subtract 6x from both sides: −15 = 1.
3. The variable vanished and what is left is false. This is a contradiction; there is no solution, ∅. Sanity check with x = 0: left = 3(−5) = −15; right = 1. They differ by 16 — and they will differ by 16 for every value of x, because both sides grow at the same rate of 6 per unit.

(c) A conditional equation. Solve 2(x + 6) = 3x + 5.

1. Distribute: 2x + 12 = 3x + 5.
2. Subtract 2x from both sides: 12 = x + 5.
3. Subtract 5 from both sides: 7 = x, that is, x = 7.
4. Check: left = 2(7 + 6) = 2(13) = 26; right = 3(7) + 5 = 21 + 5 = 26. One solution, {7}.

Why the picture explains everything

Think of each side of the equation as its own straight line. Case (c) gives the lines y = 2x + 12 and y = 3x + 5. Their slopes differ (2 versus 3), so they must cross somewhere — exactly once — and that crossing point is x = 7. Case (b) gives y = 6x − 15 and y = 6x + 1: identical slopes, different intercepts, so they are parallel and never touch. No crossing point means no solution. Case (a) gives y = 2x + 4 twice — one line lying perfectly on top of the other, touching at every single point, which is why every number is a solution.

Section 3Literal equations and formulas

Same rules, more letters

A literal equation is an equation containing more than one letter — which is to say, a formula. A = lw, d = rt, and F = (9)/(5)C + 32 are all literal equations. To "solve a formula for a variable" means to rewrite it so that your chosen letter stands alone on one side, with everything else on the other.

Here is the reassuring news: there is nothing new to learn. You use precisely the same moves from Section 1. The only difference is psychological — the other letters look like they should be doing something, and they are not. They are just numbers you happen not to know yet. If you can solve 3x = 12 by dividing by 3, you can solve bx = 12 by dividing by b.

Worked examples, from easy to genuinely tricky

(a) Solve A = ½bh for h.

1. The ½ is in the way. Multiply both sides by 2: 2A = 2 · ½bh = bh.
2. h is multiplied by b, so divide both sides by b: (2A)/(b) = h.
3. Result: h = (2A)/(b), provided b ≠ 0 (a triangle with no base has no height to find).
4. Check with real numbers: if b = 10 and h = 6, then A = ½(10)(6) = 30. Our formula gives h = (2 · 30)/(10) = (60)/(10) = 6. Correct.

(b) Solve F = (9)/(5)C + 32 for C. This is the temperature-conversion formula, and it is the classic exercise because the order of undoing matters.

1. C is multiplied by (9)/(5) and then 32 is added, so undo in reverse: subtract 32 first. F − 32 = (9)/(5)C.
2. Now undo the multiplication by (9)/(5). Multiplying by a fraction is undone by multiplying by its reciprocal, (5)/(9): (5)/(9)(F − 32) = (5)/(9) · (9)/(5)C = C.
3. Result: C = (5)/(9)(F − 32).
4. Check with a number you know: water boils at F = 212. Then C = (5)/(9)(212 − 32) = (5)/(9)(180) = 5 · 20 = 100. Water boils at 100 degrees Celsius. Correct.

(c) Solve P = 2L + 2W for W.

1. Subtract 2L from both sides: P − 2L = 2W.
2. Divide both sides by 2: W = (P − 2L)/(2).
3. Check with numbers: a rectangle with L = 9 and W = 4 has P = 2(9) + 2(4) = 18 + 8 = 26. Our formula gives W = (26 − 18)/(2) = (8)/(2) = 4. Correct.
4. A common slip: writing W = P − L. You must divide the whole numerator by 2, not just part of it.

(d) The variable appears twice: solve ax + b = cx + d for x. This one looks intimidating and is the most useful pattern in the section.

1. Gather every x-term on one side. Subtract cx from both sides: ax − cx + b = d.
2. Move the non-x terms to the other side. Subtract b: ax − cx = d − b.
3. Factor out x. This is the step people forget: x(a − c) = d − b. Factoring is what turns two x's into one.
4. Divide by the whole parenthesis. x = (d − b)/(a − c), provided a ≠ c.
5. Check against Section 1: solve 5x + 3 = 2x + 12 the long way and you get 3x = 9, x = 3. Now use the formula with a = 5, b = 3, c = 2, d = 12: x = (12 − 3)/(5 − 2) = (9)/(3) = 3. Same answer.

That restriction a ≠ c is Section 2 in disguise. If a = c, the two sides have the same slope, and the equation is either a contradiction (if b ≠ d) or an identity (if b = d). The algebra warns you by asking you to divide by zero.

Section 4Applications

The five-step method

Word problems frighten people not because the algebra is hard — it is the easiest algebra in the chapter — but because the translation step is unpractised. So do it deliberately, in writing, every time. The method never changes.

1. Read it twice and say out loud what is being asked for.
2. Name the unknown. Write "Let x = ..." with units. If there are two unknowns, define the second in terms of the first.
3. Translate the sentence into an equation, phrase by phrase.
4. Solve using Section 1.
5. Answer the question asked, in a sentence with units, and check against the original wording — not just against your own equation.

Step 5 has two halves for a reason. You can solve your equation perfectly and still be wrong if you found the brother's age when the question asked for the sister's, or if you solved for the discount when the question asked for the original price.

Worked examples: ages, money, and travel

An age problem. Maya is 4 years older than twice her brother's age. Together their ages total 40. How old is each?

1. Name the unknown. The brother's age is the simpler quantity, so let b = the brother's age in years.
2. Express the other unknown. "4 years older than twice her brother's age" means 2b + 4, so Maya's age is 2b + 4.
3. Translate. "Together their ages total 40" means b + (2b + 4) = 40.
4. Solve. Combine like terms: 3b + 4 = 40. Subtract 4: 3b = 36. Divide by 3: b = 12.
5. Answer both parts. The brother is 12. Maya is 2(12) + 4 = 24 + 4 = 28.
6. Check against the words. Twice 12 is 24, and 4 more is 28 — Maya is indeed 4 years older than twice his age. And 12 + 28 = 40. Both sentences hold.

A percent problem. A jacket sells for 68 dollars after a 15% discount. What was the original price?

1. Name the unknown. Let p = the original price in dollars. Note carefully: the discount is 15% of the original price, not of the 68 dollars.
2. Translate. Original price minus the discount equals the sale price: p − 0.15p = 68.
3. Combine like terms. p is 1p, so 1p − 0.15p = 0.85p. The equation is 0.85p = 68.
4. Solve. Divide both sides by 0.85: p = (68)/(0.85) = 80.
5. Check against the words. 15% of 80 is 0.15(80) = 12, and 80 − 12 = 68. Correct. The original price was 80 dollars.

Notice the shortcut hiding in step 3: taking 15% off is the same as keeping 85%, so you could have written 0.85p = 68 straight away. That shortcut is worth owning, because it also runs backwards for markups (a 15% increase is 1.15p).

A distance problem. Two cars leave the same rest stop at the same time and drive in opposite directions, one at 55 mi/h and the other at 65 mi/h. After how many hours are they 360 miles apart?

1. Name the unknown. Let t = the time in hours. Both cars travel for the same t.
2. Use d = rt for each car. The first covers 55t miles; the second covers 65t miles.
3. Translate. They move in opposite directions, so the gap between them is the sum of the two distances: 55t + 65t = 360.
4. Solve. Combine like terms: 120t = 360. Divide by 120: t = 3.
5. Check against the words. In 3 hours the first car goes 55(3) = 165 miles and the second goes 65(3) = 195 miles. Since they went opposite ways, they are 165 + 195 = 360 miles apart. Correct: 3 hours.

One more pattern worth seeing, because it appears constantly: three consecutive even integers sum to 84. Let n be the smallest; the next two are n + 2 and n + 4. Then n + (n + 2) + (n + 4) = 84 becomes 3n + 6 = 84, so 3n = 78 and n = 26. The integers are 26, 28, and 30, and 26 + 28 + 30 = 84. Correct.

Section 5Linear inequalities

Almost the same rules — with one exception

An inequality uses <, >, ≤, or ≥ instead of =, and asks not "which value works" but "which values work." The answer is usually a whole stretch of the number line, which is why we report it in interval notation.

You solve an inequality with exactly the moves from Section 1 — add, subtract, multiply, divide on both sides — with one rule attached: if you multiply or divide both sides by a negative number, reverse the direction of the inequality symbol. Here is why, in one line you can verify yourself. Start with the true statement −2 < 3. Multiply both sides by −1 and you get 2 and −3. Is 2 < −3? No. Is 2 > −3? Yes. Multiplying by a negative reflects both numbers across zero, which swaps which one is larger, so the symbol must flip to keep the statement true.

Note what does not trigger a flip: adding or subtracting a negative number, or having a negative number sitting somewhere in the problem. The flip happens only when the operation you perform is multiplying or dividing by a negative.

Two conventions to lock in: a square bracket means the endpoint is included (it goes with ≤ and ≥), a parenthesis means it is excluded (it goes with < and >), and ∞ always gets a parenthesis because infinity is a direction, not a number you can reach.

Here is the flip in action. Solve −3x + 7 ≤ 22 and write the answer in interval notation.

1. Subtract 7 from both sides. No flip — subtracting never flips. −3x ≤ 22 − 7, so −3x ≤ 15.
2. Divide both sides by −3. This is division by a negative, so the symbol flips from ≤ to ≥: x ≥ (15)/(−3), so x ≥ −5.
3. Write the interval. [−5, ∞) — bracket on −5 because ≥ includes it, parenthesis on infinity always.
4. Check with two test values. Try x = 0, which is in our answer: −3(0) + 7 = 7, and 7 ≤ 22 is true. Good. Now try x = −10, which is not in our answer: −3(−10) + 7 = 30 + 7 = 37, and 37 ≤ 22 is false. Correctly excluded. Finally test the endpoint x = −5: −3(−5) + 7 = 15 + 7 = 22, and 22 ≤ 22 is true, so the endpoint does belong.

Compare that with an inequality where no flip is needed. Solve 5(x − 2) > 3x + 4: distribute to get 5x − 10 > 3x + 4; subtract 3x to get 2x − 10 > 4; add 10 to get 2x > 14; divide by positive 2, so no flip, giving x > 7, or (7, ∞). Check x = 8: 5(6) = 30 and 3(8) + 4 = 28, and 30 > 28 is true.

Compound inequalities: "and" versus "or"

A compound inequality joins two inequalities. An and statement (also called a conjunction) requires both parts to be true, so its solution is the overlap — usually a single interval sandwiched between two numbers. An or statement requires at least one part to be true, so its solution is everything either part covers — usually two separate rays joined with the union symbol ∪.

An "and" example. Solve −5 ≤ 2x − 1 < 7. Because the variable sits in the middle, work on all three parts at once.

1. Add 1 to all three parts. −5 + 1 ≤ 2x − 1 + 1 < 7 + 1, giving −4 ≤ 2x < 8.
2. Divide all three parts by 2. Two is positive, so nothing flips: −2 ≤ x < 4.
3. Interval notation. [−2, 4) — closed on the left because of ≤, open on the right because of <.
4. Check. Test x = 0 (inside): 2(0) − 1 = −1, and −5 ≤ −1 < 7 is true. Test x = 4 (the excluded endpoint): 2(4) − 1 = 7, and 7 < 7 is false, so 4 is correctly left out. Test x = −2 (the included endpoint): 2(−2) − 1 = −5, and −5 ≤ −5 is true, so −2 belongs.

An "or" example. Solve 3x + 2 < −4 or 3x + 2 > 8. Solve each piece separately. From the first: 3x < −6, so x < −2. From the second: 3x > 6, so x > 2. The solution is x < −2 or x > 2, written (−∞, −2) ∪ (2, ∞). Check x = −3: 3(−3) + 2 = −7, and −7 < −4 is true, so the "or" is satisfied. Check x = 0: 3(0) + 2 = 2, which is neither less than −4 nor greater than 8, so 0 is correctly excluded.

Absolute-value inequalities

Absolute value measures distance from zero, and distance is never negative. That single fact generates both rules. If |u| < c for some positive c, then u is within c units of zero, so −c < u < c — an "and," a middle interval. If |u| > c, then u is farther than c from zero, so u < −c or u > c — an "or," two outward rays. A memory hook that has saved many students: less than is an and, greator is an or.

Worked example (less than). Solve |2x − 3| ≤ 7.

1. Rewrite without bars using the "and" form: −7 ≤ 2x − 3 ≤ 7.
2. Add 3 to all three parts. −7 + 3 ≤ 2x ≤ 7 + 3, giving −4 ≤ 2x ≤ 10.
3. Divide all three parts by 2. −2 ≤ x ≤ 5.
4. Interval notation. [−2, 5].
5. Check both endpoints. At x = 5: |2(5) − 3| = |10 − 3| = |7| = 7, and 7 ≤ 7 is true. At x = −2: |2(−2) − 3| = |−4 − 3| = |−7| = 7, also true. And at x = 6 (outside): |12 − 3| = 9, and 9 ≤ 7 is false, correctly excluded.

Worked example (greater than). Solve |3x + 1| > 5.

1. Rewrite without bars using the "or" form: 3x + 1 < −5 or 3x + 1 > 5.
2. Solve the left branch. Subtract 1: 3x < −6. Divide by 3: x < −2.
3. Solve the right branch. Subtract 1: 3x > 4. Divide by 3: x > (4)/(3).
4. Interval notation. (−∞, −2) ∪ ((4)/(3), ∞).
5. Check. Test x = 2 (in the right branch): |3(2) + 1| = |7| = 7, and 7 > 5 is true. Test x = −3 (in the left branch): |−9 + 1| = |−8| = 8, and 8 > 5 is true. Test x = 0 (in neither): |0 + 1| = 1, and 1 > 5 is false, correctly excluded.

Finally, three special cases that follow straight from "distance is never negative." |u| < −3 has no solution (∅), because no absolute value can be smaller than a negative number. |u| > −3 is true for all real numbers, (−∞, ∞), because every absolute value already beats a negative. And |u| < 0 is also empty, while |u| ≥ 0 is all reals. Recognize these by inspection rather than by grinding through the algebra — splitting them into cases produces nonsense.

ReferenceWords worth knowing

A quick reference for the vocabulary in this chapter — read it through now, then come back to it any time a word in a problem stops you cold.

Absolute value

The distance of a number from zero on the number line, written |x|; it is never negative, so |−7| = 7 and |7| = 7.

Coefficient

The number multiplying a variable. In 5x − 3, the coefficient of x is 5.

Compound inequality

Two inequalities joined by and (solution = the overlap) or or (solution = the union).

Conditional equation

An equation true only for particular value(s) of the variable — the ordinary case, with exactly one solution for a linear equation.

Constant

A term with no variable in it, such as the −3 in 5x − 3.

Contradiction

An equation true for no value of the variable; solving it produces a false statement such as 4 = 9, and the solution set is empty (∅).

Distributive property

The rule a(b + c) = ab + ac, used to clear parentheses.

Equivalent equations

Two equations with exactly the same solution set; every legal move you make produces one.

Extraneous solution

A value produced by correct algebra that nevertheless fails in the original equation — typically because it violates a restriction.

Identity

An equation true for every allowed value of the variable; solving it produces a true statement such as 6 = 6, and the solution set is all real numbers.

Inequality

A statement comparing two expressions with <, >, ≤, or ≥ rather than =.

Interval notation

A compact way to write a solution set: brackets [ ] include an endpoint, parentheses ( ) exclude it, and ∞ always takes a parenthesis.

Least common denominator (LCD)

The smallest expression every denominator divides into; multiplying an equation by it clears all fractions at once.

Linear equation

An equation in which the variable appears only to the first power, with no variable under a radical or alone in a denominator.

Literal equation

An equation containing two or more letters — a formula — which can be solved for any one of them.

No solution (empty set)

The solution set of a contradiction, written ∅ or { }; it is not the same as the solution x = 0.

Properties of equality

The rules permitting you to add, subtract, multiply, or divide both sides of an equation by the same quantity (nonzero for multiplying and dividing).

Restriction (excluded value)

A value the variable may not take because it would make a denominator zero or otherwise undefine the expression.

Solution (root)

A value that makes an equation a true statement when substituted for the variable.

Solution set

The collection of all solutions, written in set or interval notation.

Term

A single number, variable, or product of them, separated from its neighbours by + or − signs. In 5x − 3 there are two terms.

Variable

A letter standing for an unknown or changing quantity.

Graphs & Lines

Algebra stops being a pile of symbols the moment you can see it. A graph turns an equation into a picture, and the simplest, most useful picture in all of mathematics is a straight line. In this chapter you will learn to place a point on the plane, measure how steeply a line climbs or falls, write that line down in three interchangeable ways, recognize when two lines run side by side or cross at a perfect corner, and finally use a line to describe something real — a price, a distance, a rate. Every idea here is built from arithmetic you already own: subtraction, division, and the willingness to check your answer. If you have ever felt lost in a math class, this is a good place to get found, because almost everything is verifiable — you can plug your answer back in and watch it work.

Section 1The coordinate plane

Two number lines, one address system

Take the number line you have known since grade school and lay it flat. Now take a second number line and stand it up, crossing the first at zero. That is the whole invention. The horizontal line is the x-axis, the vertical line is the y-axis, and the crossing point is the origin, written (0, 0). Together they form the Cartesian coordinate plane, named for René Descartes, and their only job is to give every point in the flat plane a unique two-number address.

That address is an ordered pair, written (x, y). The word ordered is doing real work: the first number is always the horizontal move and the second is always the vertical move. To plot (4, −3), start at the origin, walk 4 to the right because 4 is positive, then walk 3 down because −3 is negative, and put your dot there. To plot (−3, 4), start at the origin, walk 3 to the left, then 4 up. Those are two completely different points, which is exactly why the order matters.

A useful habit: say the moves out loud as you make them. "Four right, three down." Students who narrate the two moves make far fewer sign mistakes than students who try to eyeball it.

The four quadrants

The two axes chop the plane into four regions called quadrants, numbered with Roman numerals starting in the upper right and going counterclockwise. That counterclockwise direction is arbitrary but universal, so it is worth memorizing once. The quadrant a point lands in is determined entirely by the two signs in its address — you never have to look at a picture to know the answer.

Notice that last row. A point sitting on an axis is not in any quadrant — the axes are the fences between the yards, not part of any yard. So (0, 5) is on the y-axis, (−4, 0) is on the x-axis, and the origin is on both. Test questions love this.

Intercepts: where a graph meets the axes

An intercept is a point where a graph crosses an axis, and the two kinds are defined by a single fact each. The x-intercept is where the graph crosses the x-axis, and every point on the x-axis has y = 0. The y-intercept is where the graph crosses the y-axis, and every point on the y-axis has x = 0. That gives you a recipe with no memorization required:

• To find the x-intercept, set y = 0 and solve for x.
• To find the y-intercept, set x = 0 and solve for y.

Fully worked example. Find both intercepts of 2x + 5y = 20, and use them to graph the line.

Step 1 — find the x-intercept by setting y = 0.

• 2x + 5y = 20  →  2x + 5(0) = 20
• 5(0) = 0, so the equation becomes 2x + 0 = 20, that is 2x = 20
• Divide both sides by 2: (2x)/(2) = (20)/(2), so x = 10
• The x-intercept is the point (10, 0).

Step 2 — find the y-intercept by setting x = 0.

• 2(0) + 5y = 20
• 2(0) = 0, so 0 + 5y = 20, that is 5y = 20
• Divide both sides by 5: (5y)/(5) = (20)/(5), so y = 4
• The y-intercept is the point (0, 4).

Step 3 — check both answers by substituting them back into the original equation.

• Check (10, 0): 2(10) + 5(0) = 20 + 0 = 20.  True. ✓
• Check (0, 4): 2(0) + 5(4) = 0 + 20 = 20.  True. ✓

Step 4 — graph. Plot (10, 0) and (0, 4), lay a straightedge across them, and draw. Two points determine a line, so an intercept pair is often the fastest possible graph. A third point is a cheap insurance policy: try x = 5, which gives 2(5) + 5y = 20 → 10 + 5y = 20 → 5y = 10 → y = 2, so (5, 2) should land right on your line — and it does, exactly halfway between the intercepts.

Section 2Slope

What slope actually measures

Slope answers one question: when I move to the right, how fast does the line go up or down? It is a ratio — vertical change divided by horizontal change — and it is usually called rise over run. Rise is how far you climbed (negative if you fell). Run is how far you traveled sideways. Slope is written with the letter m, a convention with no good explanation that we are all stuck with.

Given any two distinct points (x₁, y₁) and (x₂, y₂) on a line:

m = (rise)/(run) = (y₂ − y₁)/(x₂ − x₁),   provided x₂ ≠ x₁

That restriction — x₂ ≠ x₁ — is not decoration. If the two x-values are equal, the denominator is zero, and division by zero is undefined. We will see in a moment exactly what kind of line does that.

Two comforts. First, it does not matter which point you call "point 1." Swap them and both the top and the bottom flip sign, and the two negatives cancel. Second, it does not matter which two points on the line you pick — a straight line has the same steepness everywhere, which is precisely what makes it straight. Slope is the line's one defining number.

A fully worked slope calculation

Fully worked example. Find the slope of the line through (−3, 7) and (5, −1).

Step 1 — label the points so the substitution is mechanical.

• x₁ = −3,  y₁ = 7  (the first point)
• x₂ = 5,  y₂ = −1  (the second point)

Step 2 — substitute into the formula, writing the parentheses even when they look silly.

• m = (y₂ − y₁)/(x₂ − x₁) = (−1 − 7)/(5 − (−3))

Step 3 — do the top and the bottom separately, out loud.

• Top: −1 − 7 = −8.  (Starting at −1 and subtracting 7 moves you further left on the number line.)
• Bottom: 5 − (−3) = 5 + 3 = 8.  (Subtracting a negative is adding — this is the single most common place to slip.)

Step 4 — divide.

• m = (−8)/(8) = −1

Step 5 — check by reversing the labels. Call (5, −1) the first point instead:

• m = (7 − (−1))/(−3 − 5) = (7 + 1)/(−8) = (8)/(−8) = −1.  Same answer. ✓

What it means. A slope of −1 says: every 1 unit you move right, the line drops 1 unit. It falls at a 45° angle.

One more, with a fraction that does not simplify to a whole number. Through (2, −5) and (6, 1):

• m = (1 − (−5))/(6 − 2) = (1 + 5)/(4) = (6)/(4) = (3)/(2)

Always reduce. A slope of (3)/(2) is a clean instruction: run 2 right, rise 3 up. Left as 6/4 it still works — run 4, rise 6 lands on the same line — but reduced is the standard form and the one graders expect.

Four kinds of slope

Every line falls into exactly one of four categories, and you should be able to name the category before you finish the arithmetic.

Study the last two rows side by side, because they are the pair that gets confused. Zero slope means the rise is zero — you moved sideways and never went up. That is a real number, zero, and the line is horizontal with equation y = (some number). Undefined slope means the run is zero — you went up without going sideways at all — and there is no number that equals 6 divided by 0. The line is vertical with equation x = (some number). "Zero" and "no slope" are not the same thing, and saying "no slope" out loud is how the confusion starts, so prefer the word undefined.

Slope as a rate of change

Outside of a graph, slope has a much more familiar name: a rate. Miles per hour, dollars per page, degrees per minute, milligrams per kilogram — every "per" you have ever said is a slope. The units tell you exactly what the number means: slope has units of (y-units) per (x-unit). If x is measured in hours and y in miles, a slope of 55 is 55 miles per hour. If x is pages and y is dollars, a slope of 0.18 is 18 cents per page. This is the bridge from Section 2 to Section 5, and it is the reason lines matter at all.

Section 3Forms of a line

Three outfits, one line

A single line can be written in several algebraically equivalent ways. They describe the same set of points; they just make different information easy to read at a glance. Learning all three is not busywork — each one is the fastest tool for a particular job.

In slope-intercept form, y = mx + b, the number multiplying x is the slope and the number standing alone is the y-value where the line crosses the y-axis. In y = −4x + 9, the slope is −4 and the y-intercept is (0, 9) — no work required. To graph it: put a dot at (0, 9), then use the slope −4 = (−4)/(1) to step 1 right and 4 down to (1, 5), and connect.

In standard form, Ax + By = C, both variables sit on the left and the constant sits alone on the right. By convention A, B, and C are integers with no common factor and A is non-negative. Standard form is where the intercept recipe from Section 1 shines, and it is the form vertical lines can live in (x = 2 is 1x + 0y = 2) even though slope-intercept form cannot hold them.

Point-slope: the workhorse

Point-slope form deserves special affection because it is the form you build with. Rearrange the slope formula: if (x, y) is any point on the line and (x₁, y₁) is a known point, then m = (y − y₁)/(x − x₁). Multiply both sides by (x − x₁) and you get y − y₁ = m(x − x₁). That is it — point-slope form is just the slope formula with the denominator cleared. Whenever a problem gives you a point and a slope, or two points, start here.

Fully worked example. Write the equation of the line through (−2, 5) with slope (−3)/(4). Give it in point-slope, slope-intercept, and standard form.

Step 1 — write point-slope, substituting carefully. Here x₁ = −2, y₁ = 5, m = (−3)/(4).

• y − y₁ = m(x − x₁)
• y − 5 = (−3)/(4) · (x − (−2))
• Simplify the inside: x − (−2) = x + 2
• y − 5 = (−3)/(4) · (x + 2)   ← point-slope form, done

Step 2 — distribute to reach slope-intercept form.

• (−3)/(4) · x = (−3x)/(4)
• (−3)/(4) · 2 = (−6)/(4) = (−3)/(2)
• So y − 5 = (−3x)/(4) − (3)/(2)

Step 3 — add 5 to both sides and combine the constants over a common denominator.

• y = (−3x)/(4) − (3)/(2) + 5
• Rewrite 5 as (10)/(2) so the fractions match: −(3)/(2) + (10)/(2) = (7)/(2)
• y = (−3)/(4) · x + (7)/(2)   ← slope-intercept form. Slope (−3)/(4), y-intercept (0, (7)/(2)) = (0, 3.5)

Step 4 — clear the fractions to reach standard form. The denominators are 4 and 2, so multiply every term by 4.

• 4 · y = 4 · (−3x)/(4) + 4 · (7)/(2)
• 4y = −3x + 14
• Add 3x to both sides so A is positive: 3x + 4y = 14
• 3x + 4y = 14   ← standard form (integers, no common factor, A > 0)

Step 5 — check the original point in all three forms. Substitute x = −2 and see whether y comes out 5.

• Point-slope: y − 5 = (−3)/(4) · (−2 + 2) = (−3)/(4) · 0 = 0, so y = 5. ✓
• Slope-intercept: y = (−3)/(4) · (−2) + (7)/(2) = (6)/(4) + (7)/(2) = (3)/(2) + (7)/(2) = (10)/(2) = 5. ✓
• Standard: 3(−2) + 4(5) = −6 + 20 = 14. ✓

Three forms, one line, all verified.

Two points to an equation

If you are handed two points instead of a slope, you simply do Section 2 first and then Section 3. Nothing new is required.

Fully worked example. Find the equation of the line through (1, −2) and (4, 7).

• Step 1 — slope. m = (7 − (−2))/(4 − 1) = (7 + 2)/(3) = (9)/(3) = 3
• Step 2 — point-slope, using (1, −2). y − (−2) = 3(x − 1), which tidies to y + 2 = 3(x − 1)
• Step 3 — distribute. y + 2 = 3x − 3
• Step 4 — subtract 2 from both sides. y = 3x − 3 − 2 = 3x − 5
• Step 5 — check the point you did NOT use. At x = 4: y = 3(4) − 5 = 12 − 5 = 7. ✓ The line passes through (4, 7) as required.

Notice Step 5. Using the unused point as your check is free insurance: if the arithmetic had gone wrong anywhere, that substitution would have caught it. Had we started with (4, 7) instead, we would have gotten y − 7 = 3(x − 4) → y = 3x − 12 + 7 = 3x − 5 — the identical line. Either point works.

Reading slope out of standard form

Given 5x − 2y = 8, you cannot see the slope. Solve for y and it appears. One condition first: solving Ax + By = C for y requires dividing by B, so this move is legal only when B ≠ 0. If B = 0 the equation is Ax = C, a vertical line, which has no slope-intercept form at all. Here B = −2, so we are safe:

• Subtract 5x: −2y = −5x + 8
• Divide every term by −2: (−2y)/(−2) = (−5x)/(−2) + (8)/(−2)
• y = (5)/(2) · x − 4
• So m = (5)/(2) and the y-intercept is (0, −4).

The dangerous step is dividing by a negative: both the x-term and the constant change sign. Divide term by term, in writing, every time.

Section 4Parallel and perpendicular lines

Parallel means equal slopes

Two lines are parallel when they never intersect, no matter how far you extend them. Since slope is the one number that controls direction, this happens exactly when the slopes match: m₁ = m₂. One caution — equal slopes and equal y-intercepts do not give you two parallel lines; they give you the same line drawn twice. For genuinely parallel lines you need equal slopes and different intercepts.

Perpendicular means negative reciprocal slopes

Two lines are perpendicular when they cross at a right angle (90°). The slope condition is:

m₁ · m₂ = −1,  equivalently  m₂ = −1/m₁  (provided m₁ ≠ 0)

The condition m₁ ≠ 0 on that second version is the same excluded-value habit from the slope formula: the reciprocal form divides by m₁, so it says nothing at all when m₁ = 0. That gap is exactly the horizontal-line case treated below.

In plain language: flip the fraction and change the sign. That is what "negative reciprocal" means, and doing it in two deliberate motions prevents the classic half-error of flipping without negating. If m₁ = (2)/(3), flip to (3)/(2), then negate to get (−3)/(2). If m₁ = −5, first write it as (−5)/(1), flip to (1)/(−5) = −(1)/(5), then negate to get (1)/(5). Confirm: (−5) · (1)/(5) = (−5)/(5) = −1. ✓

There is one honest exception. A horizontal line (m = 0) and a vertical line (slope undefined) clearly meet at a right angle, but you cannot multiply 0 by "undefined" and get −1. The product rule simply does not apply to that pair — they are perpendicular by geometry, not by arithmetic. Handle vertical and horizontal lines by picture, not by formula.

A fully worked pair

Fully worked example. Line L has equation 2x − 3y = 12. Find (a) the line parallel to L through (6, −1) and (b) the line perpendicular to L through (6, −1). Give both in slope-intercept form.

Step 1 — get the slope of L by solving for y.

• 2x − 3y = 12
• Subtract 2x: −3y = −2x + 12
• Divide every term by −3: (−3y)/(−3) = (−2x)/(−3) + (12)/(−3)
• y = (2)/(3) · x − 4,  so m_L = (2)/(3)

Step 2(a) — the parallel line has the same slope, (2)/(3). Use point-slope with (6, −1):

• y − (−1) = (2)/(3) · (x − 6), that is y + 1 = (2)/(3) · (x − 6)
• Distribute: (2)/(3) · 6 = (12)/(3) = 4, so y + 1 = (2)/(3) · x − 4
• Subtract 1: y = (2)/(3) · x − 5
• Check the point: at x = 6, y = (2)/(3) · 6 − 5 = 4 − 5 = −1. ✓
• Check it is truly parallel: slope (2)/(3) matches L, and the intercept −5 differs from L's −4, so it is a different line. ✓

Step 2(b) — the perpendicular line uses the negative reciprocal. Flip (2)/(3) to (3)/(2), then negate: m_⊥ = (−3)/(2).

• y + 1 = (−3)/(2) · (x − 6)
• Distribute: (−3)/(2) · (−6) = (18)/(2) = 9, so y + 1 = (−3)/(2) · x + 9
• Subtract 1: y = (−3)/(2) · x + 8
• Check the point: at x = 6, y = (−3)/(2) · 6 + 8 = −9 + 8 = −1. ✓
• Check the right angle: (2)/(3) · (−3)/(2) = (−6)/(6) = −1. ✓

Deciding a relationship. To classify two given lines, put both into slope-intercept form and compare. Are 3x − 4y = 20 and y = (3)/(4) · x − 2 parallel? Solve the first: −4y = −3x + 20 → y = (3)/(4) · x − 5. Both slopes are (3)/(4); the intercepts −5 and −2 differ. Parallel.

Section 5Modeling with linear equations

When is a line the right model?

A linear model is appropriate whenever a quantity changes by the same amount in every equal step: the same dollars per page, the same gallons per minute, the same degrees per hour. That constant step is the slope. If instead a quantity keeps multiplying — doubling, or growing by a fixed percentage — a line is the wrong tool and you need the curves of later chapters. Before you fit a line, ask: is the change constant, or is it compounding?

Two habits separate a model from a bare equation. First, name your variables with units: not "x" but "p = number of pages printed." Second, say the slope and intercept in a sentence. A model that you cannot explain in English has not been finished.

Building a model from two data points

Fully worked example. A print shop charges a one-time setup fee plus a fixed cost per page. A 100-page order costs $23. A 400-page order costs $77. Write a linear model for the cost, interpret it, and use it to price a 250-page order.

Step 1 — define the variables, with units.

• p = number of pages printed (the input, so it goes on the horizontal axis)
• C = total cost in dollars (the output)
• The two facts become the ordered pairs (100, 23) and (400, 77).

Step 2 — find the slope, which is the cost per page.

• m = (77 − 23)/(400 − 100)
• Top: 77 − 23 = 54 dollars
• Bottom: 400 − 100 = 300 pages
• m = (54)/(300) = (54 ÷ 6)/(300 ÷ 6) = (9)/(50) = 0.18
• Units: dollars per page. So each page costs $0.18, or 18 cents.

Step 3 — use point-slope with either data point. We will use (100, 23).

• C − 23 = 0.18(p − 100)
• Distribute: 0.18 · 100 = 18, so C − 23 = 0.18p − 18
• Add 23 to both sides: C = 0.18p − 18 + 23
• −18 + 23 = 5
• C = 0.18p + 5

Step 4 — check with the data point we did not use.

• At p = 400: C = 0.18(400) + 5 = 72 + 5 = 77.  Matches the given $77. ✓

Step 5 — interpret in plain English (this is the actual point of the exercise).

• Slope 0.18: every additional page adds $0.18 to the bill. Cost rises 18 cents per page.
• y-intercept 5: at p = 0 pages the cost is $5. That is the setup fee — what you pay just to walk in the door before a single page prints.

Step 6 — use the model to predict. Price a 250-page order:

• C = 0.18(250) + 5
• 0.18 · 250 = 45
• C = 45 + 5 = $50.00

Step 7 — state the restriction. Pages cannot be negative, so the model is only meaningful for p ≥ 0, and in practice p should be a whole number. Plugging in p = −20 produces a number, but not a meaning.

Reading a model that is handed to you

Half of modeling is interpretation, not construction. Suppose a tank is draining and the volume of water is V = 500 − 12t, where V is gallons and t is minutes. Rewrite it mentally as V = −12t + 500 so the shape is familiar.

• Slope −12: the tank loses 12 gallons every minute. Negative slope means the quantity is decreasing — the sign carries meaning.
• Intercept 500: at t = 0, V = −12(0) + 500 = 500 gallons. The tank started full at 500 gallons.
• When is it empty? Set V = 0:  0 = 500 − 12t → 12t = 500 → t = (500)/(12) = (125)/(3) ≈ 41.7 minutes.
• Check: at t = (125)/(3), V = 500 − 12 · (125)/(3) = 500 − (1500)/(3) = 500 − 500 = 0. ✓
• Restriction: the model is meaningful only for 0 ≤ t ≤ (125)/(3). After that the tank is empty; the equation would report negative gallons, which is nonsense.

That last bullet is worth a moment. Predicting inside the range of your data is interpolation and is usually safe. Predicting outside it is extrapolation, and it is where linear models go to embarrass people. Our print-shop model happily reports a price for a one-million-page order; the actual shop would give you a bulk discount, and the line would be wrong. A model is a description of a range, not a law of nature.

Average rate of change

When data are not perfectly linear, the slope between two data points still has a name: the average rate of change over that interval. If a cooling cup of coffee measures 186°F at minute 2 and 174°F at minute 8, the average rate of change is (174 − 186)/(8 − 2) = (−12)/(6) = −2 degrees per minute. The coffee did not necessarily drop exactly 2 degrees in every single minute — but on average, over that span, that is the trend. Same formula, humbler claim.

ReferenceWords worth knowing

A quick reference for the vocabulary in this chapter — skim it now, and circle back whenever a word slows you down mid-problem.

Average rate of change

The slope between two data points, (y₂ − y₁)/(x₂ − x₁), describing the typical change per unit over that interval even when the data are not perfectly linear.

Cartesian coordinate system

The grid formed by a horizontal x-axis and a vertical y-axis crossing at the origin, giving every point an address (x, y).

Collinear

Lying on the same straight line. Three points are collinear when the slope between each pair of them is the same — or when all three share the same x-value, the vertical case, where every pairwise slope is undefined.

Extrapolation

Using a model to predict outside the range of the data it was built from — possible, but unreliable, and the usual source of absurd predictions.

Horizontal line

A flat line with slope 0 and equation y = k; every point on it shares the same y-value.

Intercept

A point where a graph crosses an axis. Always reported as an ordered pair, never as a lone number.

Interpolation

Using a model to estimate a value inside the range of the data it was built from — generally the safer kind of prediction.

Linear equation in two variables

An equation whose graph is a straight line; it can always be written as Ax + By = C with A and B not both zero.

Negative reciprocal

The result of flipping a fraction and changing its sign; the negative reciprocal of (2)/(3) is (−3)/(2). Perpendicular slopes are negative reciprocals of each other.

Ordered pair

A point written (x, y), where the first entry is the horizontal position and the second is the vertical position.

Origin

The point (0, 0), where the two axes cross.

Parallel lines

Lines that never intersect; they have equal slopes and different y-intercepts.

Perpendicular lines

Lines that cross at a 90° angle; their slopes multiply to −1 (with horizontal-and-vertical pairs as the geometric exception).

Point-slope form

y − y₁ = m(x − x₁) — the form built directly from one known point and the slope.

Quadrant

One of the four regions the axes divide the plane into, numbered I through IV counterclockwise from the upper right. Points on an axis are in no quadrant.

Rise and run

The vertical change (rise) and horizontal change (run) between two points; their quotient is the slope.

Slope

The constant rate at which a line rises or falls, m = (y₂ − y₁)/(x₂ − x₁), defined only when x₂ ≠ x₁.

Slope-intercept form

y = mx + b, where m is the slope and (0, b) is the y-intercept.

Standard form of a line

Ax + By = C, conventionally with integer coefficients, no common factor, and A non-negative.

Undefined slope

What happens when the run is zero and the slope formula would divide by zero; the line is vertical. Not the same as zero slope.

x-intercept

The point where a graph crosses the x-axis, found by setting y = 0 and solving; it has the form (a, 0).

y-intercept

The point where a graph crosses the y-axis, found by setting x = 0 and solving; it has the form (0, b).

Introduction to Functions

Up to now, algebra has mostly been about solving — find the x that makes the sentence true. This chapter changes the question. Instead of hunting for one answer, we start studying relationships: how one quantity depends on another, all at once, for every possible input. That idea is called a function, and it is the single most important concept in the rest of your math life. Nothing here requires new arithmetic — if you can substitute a number into an expression and simplify, you already have every skill you need. What changes is the way we talk about it, and this chapter walks that new language one careful step at a time.

Section 1What a function is

The definition, in plain words

A relation is any pairing of inputs with outputs. That is a loose idea — a relation can pair anything with anything. A function is a relation with one extra rule attached, and the rule is short enough to memorize today:

A function assigns to each input exactly one output.

Read that slowly, because every word is load-bearing. Each input must get an output. And each input gets exactly one — not two, not "it depends." Think of a vending machine: you press B4 and you get one specific snack. If pressing B4 sometimes gave you chips and sometimes gave you a candy bar, the machine would be broken. A function is a machine that is not broken.

Notice what the rule does not say. It says nothing about outputs being used only once. Two different inputs are perfectly allowed to give the same output — press B4 and B5 and get the same brand of chips, no problem. The restriction runs in one direction only: one input, one output. Students lose points on this more than almost anything else in the chapter, so it is worth writing on a sticky note.

The input is called the independent variable (usually x), because you get to choose it freely. The output is the dependent variable (usually y), because its value depends on what you chose.

Five ways to spot a function

Relations show up in five common outfits: a list of ordered pairs, a table, a mapping diagram (arrows from a left-hand set to a right-hand set), a graph, and an equation. The function test looks slightly different in each costume, but it is always the same question underneath: does any input have two different outputs?

Worked example 1 — ordered pairs

Decide whether each relation is a function.

Relation A: {(1, 3), (2, 5), (3, 7), (4, 9)}

Step 1. List the inputs (first coordinates): 1, 2, 3, 4.
Step 2. Ask if any repeats. They do not — each input appears exactly once.
Step 3. Conclusion: A is a function.

Relation B: {(1, 3), (2, 5), (3, 7), (2, 9)}

Step 1. List the inputs: 1, 2, 3, 2.
Step 2. The input 2 appears twice. Check its outputs: the pair (2, 5) says the output is 5, and the pair (2, 9) says the output is 9. Since 5 ≠ 9, the input 2 has two different outputs.
Step 3. Conclusion: B is not a function.

Relation C: {(1, 4), (2, 4), (3, 4), (4, 4)}

Step 1. Inputs: 1, 2, 3, 4 — no repeats.
Step 2. The outputs are all 4, which looks suspicious but breaks no rule. Repeated outputs are legal.
Step 3. Conclusion: C is a function (a constant function, in fact).

Worked example 2 — the vertical line test

Does the equation y² = x make y a function of x?

Step 1. Pick a specific input and see what happens. Let x = 9.
Step 2. Substitute: y² = 9.
Step 3. Solve. Taking a square root of both sides gives y = ±3, because 3² = 9 and (−3)² = 9. Both are genuine solutions.
Step 4. Interpret. The single input x = 9 produced two outputs, y = 3 and y = −3. That violates the one-input-one-output rule.
Step 5. Conclusion: y² = x is not a function of x.

Now look at it graphically. The graph of y² = x is a parabola opening sideways to the right. Draw the vertical line x = 9. It pierces the curve at the point (9, 3) and again at (9, −3) — two hits. That is exactly the vertical line test: if any vertical line crosses the graph more than once, the graph is not a function.

The vertical line test is not a separate rule you have to memorize on top of the definition — it is the definition, drawn. A vertical line is the set of all points sharing one x-value. If the line hits the graph twice, you have found one x with two y's. That is the whole argument.

Section 2Function notation

What f(x) means — and what it does not

Instead of writing y = 3x + 1, mathematicians usually write f(x) = 3x + 1. Out loud, that is "f of x equals three x plus one." The letter f is the name of the function, like the name of a recipe. The x in parentheses is the input slot.

Here is the point of confusion that trips up nearly everyone, so let us kill it immediately: f(x) does not mean f times x. The parentheses are not multiplication parentheses. They are a holder announcing "whatever goes in here is the input." If you see f(3), it does not mean 3f — it means "run the number 3 through the rule named f."

Why bother with the extra notation at all? Because it is precise and compact. The single symbol f(3) says "the output of this function when the input is 3." Saying that in y-language takes a whole sentence. And once you have several functions in one problem — f, g, h, or C(x) for cost and R(x) for revenue — the naming keeps them straight.

Worked example 3 — evaluating at a number

Let f(x) = 3x² − 5x + 2. Find f(−2).

Step 1. Rewrite the rule with an empty slot where every x was:
f( ) = 3( )² − 5( ) + 2

Step 2. Drop −2 into every slot. Always use parentheses — this is the step that saves you from sign errors:
f(−2) = 3(−2)² − 5(−2) + 2

Step 3. Exponents first. (−2)² = (−2)(−2) = 4. Note the parentheses matter enormously here: (−2)² = 4, but −2² = −4.
f(−2) = 3(4) − 5(−2) + 2

Step 4. Multiply. 3(4) = 12, and −5(−2) = +10 (negative times negative is positive):
f(−2) = 12 + 10 + 2

Step 5. Add left to right. 12 + 10 = 22, then 22 + 2 = 24.
f(−2) = 24

What did we actually learn? That the point (−2, 24) sits on the graph of f. Every evaluation is a point.

Worked example 4 — evaluating at an expression

Let f(x) = x² − 3. Find f(a + 1).

Step 1. Every x becomes the entire quantity (a + 1), parentheses included:
f(a + 1) = (a + 1)² − 3

Step 2. Expand the square. Remember (a + 1)² means (a + 1)(a + 1), not a² + 1:
(a + 1)(a + 1) = a² + a + a + 1 = a² + 2a + 1

Step 3. Substitute back and combine:
f(a + 1) = a² + 2a + 1 − 3 = a² + 2a − 2

Step 4. Sanity check with a number. Let a = 3. Our answer predicts 3² + 2(3) − 2 = 9 + 6 − 2 = 13. Directly: f(3 + 1) = f(4) = 4² − 3 = 16 − 3 = 13. They match, so the algebra is right.

Worked example 5 — going backwards

Let f(x) = 2x + 7. Find the value of x for which f(x) = 19.

Step 1. Notice the direction. Here we are handed the output and asked for the input — the reverse of example 3.
Step 2. Set the rule equal to the given output: 2x + 7 = 19.
Step 3. Subtract 7 from both sides: 2x + 7 − 7 = 19 − 7, so 2x = 12.
Step 4. Divide both sides by 2: x = 6.
Step 5. Check by substituting back: f(6) = 2(6) + 7 = 12 + 7 = 19. ✓
x = 6

Section 3Domain and range

Two sets, one function

The domain is the set of every input the function is allowed to accept. The range is the set of every output it actually produces. If a function is a machine, the domain is the list of raw materials it can process and the range is the list of products that come out.

Most algebra functions accept every real number, and when that happens we simply say the domain is all real numbers. Trouble only arrives from two places, and in this course those two are the entire list:

• Division by zero is undefined. Any input that makes a denominator equal zero must be thrown out.
• Even roots of negative numbers are not real. Any input that makes the inside of a square root (or fourth root, sixth root, …) negative must be thrown out.

That is it. If the function has no fraction with a variable underneath and no square root with a variable inside, its domain is all real numbers — you do not have to hunt for a problem that is not there.

Interval notation

Domains and ranges are usually written in interval notation, a compact shorthand. A square bracket means "include this endpoint," a parenthesis means "get as close as you like but do not include it." Infinity is not a number you can reach, so it always gets a parenthesis. The union symbol ∪ means "and also this piece."

Worked example 6 — domain of a rational function

Find the domain of f(x) = (x + 2)/(x² − 9).

Step 1. Spot the hazard. There is a variable in the denominator, so division by zero is possible. There is no square root, so that hazard does not apply.
Step 2. Set the denominator equal to zero and solve — these are the values to exclude:
x² − 9 = 0
Step 3. Add 9 to both sides: x² = 9.
Step 4. Take the square root of both sides, keeping both signs: x = ±3, so x = 3 or x = −3.
Step 5. Verify one of them. At x = 3 the denominator is 3² − 9 = 9 − 9 = 0, and the numerator is 3 + 2 = 5, giving 5/0, which is undefined. Confirmed.
Step 6. Write the answer. The domain is every real number except 3 and −3:
(−∞, −3) ∪ (−3, 3) ∪ (3, ∞)

Notice we did not touch the numerator. The numerator being zero is perfectly fine — f(−2) = 0/((−2)² − 9) = 0/(4 − 9) = 0/(−5) = 0, a legitimate output. Only zero in the denominator is fatal.

Worked example 7 — domain and range of a square root function

Find the domain and range of g(x) = √(2x − 8).

Step 1. Spot the hazard. A square root, so the inside must not be negative.
Step 2. Set the inside greater than or equal to zero:
2x − 8 ≥ 0
Step 3. Add 8 to both sides: 2x ≥ 8.
Step 4. Divide both sides by 2. Dividing by a positive number leaves the inequality direction alone: x ≥ 4.
Step 5. Domain in interval notation: [4, ∞). The bracket on 4 is correct because x = 4 is allowed — it gives √0, and √0 = 0 is a real number.
Step 6. Test the boundary and one point on each side. g(4) = √(2(4) − 8) = √(8 − 8) = √0 = 0 ✓. g(6) = √(2(6) − 8) = √(12 − 8) = √4 = 2 ✓. g(3) = √(2(3) − 8) = √(6 − 8) = √(−2), which is not a real number — correctly excluded.
Step 7. Range. The principal square root symbol never returns a negative value, so the smallest possible output is 0 (reached at x = 4), and outputs grow without bound as x grows. Range: [0, ∞).

Worked example 8 — both hazards at once, and reading a graph

Find the domain of h(x) = 5/√(x − 1).

Step 1. There is a square root and that root is in a denominator, so both rules apply.
Step 2. The root requires x − 1 ≥ 0, that is x ≥ 1.
Step 3. The denominator requires √(x − 1) ≠ 0, which fails only when x − 1 = 0, that is x = 1.
Step 4. Combine: x ≥ 1 but x ≠ 1 means x > 1. Domain: (1, ∞). The endpoint that was included in step 2 gets kicked back out in step 3 — this is exactly why we check both rules.

Reading a graph is easier still: domain is what you see scanning left to right, range is what you see scanning bottom to top. For the parabola y = x² − 4, the curve stretches forever left and forever right, so the domain is (−∞, ∞). Vertically, the lowest point is the vertex at (0, −4) — check it: y = 0² − 4 = −4 — and the arms rise forever, so the range is [−4, ∞), with a bracket because −4 is actually attained.

Section 4Graphs of basic functions

The toolkit

There are six graphs that show up constantly in college algebra, and most of the functions you meet in the next several chapters are one of these six wearing a disguise. (Two more parents — the exponential and the logarithmic — join the family later; no amount of shifting or stretching a toolkit graph produces 2^x.) They are called the toolkit functions (or parent functions). Learning their shapes now means that later, when you meet y = −2(x − 3)² + 5, you will not start from scratch — you will recognize a parabola and adjust.

You do not memorize these by staring at pictures. You memorize them by plotting three or four points by hand, once, and letting the shape emerge. Do that today and it sticks.

Worked example 9 — building the cubic by hand

Make a table of values for f(x) = x³ using x = −2, −1, 0, 1, 2, and describe the shape.

f(−2) = (−2)³ = (−2)(−2)(−2). Work left to right: (−2)(−2) = 4, then 4(−2) = −8. So f(−2) = −8.
f(−1) = (−1)³ = (−1)(−1)(−1) = 1(−1) = −1.
f(0) = 0³ = 0.
f(1) = 1³ = 1.
f(2) = 2³ = (2)(2)(2) = 4(2) = 8.

Plot (−2, −8), (−1, −1), (0, 0), (1, 1), (2, 8) and connect smoothly. You get the S-curve: it comes up steeply from the lower left, flattens briefly at the origin, then climbs steeply again. Because cubing keeps the sign of the input and the outputs grow without bound in both directions — down toward −∞ on the left and up toward ∞ on the right, with no gaps along the way — the cubic's range is all real numbers — a genuine difference from the quadratic in the row above it, which never goes below 0.

Compare the three rows in that table and the differences jump out. The squaring row and the absolute-value row are both symmetric — the value at −2 equals the value at 2 — which is why both graphs fold neatly over the y-axis. The cubing row is anti-symmetric: −8 versus 8. That is why the cubic never folds; it spins.

Worked example 10 — square root and reciprocal

For f(x) = √x, choose perfect squares so the arithmetic stays clean:
f(0) = √0 = 0, because 0² = 0.
f(1) = √1 = 1, because 1² = 1.
f(4) = √4 = 2, because 2² = 4.
f(9) = √9 = 3, because 3² = 9.
Points (0, 0), (1, 1), (4, 2), (9, 3) show the signature behavior: the curve rises quickly at first, then flattens. Nothing exists to the left of x = 0, which is the domain [0, ∞) drawn.

For f(x) = 1/x:
f(−2) = 1/(−2) = −½.
f(−½) = 1/(−½) = −2, since dividing by one-half doubles.
f(0) is undefined — 1/0 has no value.
f(½) = 1/(½) = 2.
f(2) = 1/2 = ½.
Two things follow. As x approaches 0 the outputs blow up in size, so the graph races up the y-axis on the right and down it on the left without ever touching it. And as x grows large the outputs shrink toward 0 without reaching it. Those two invisible guide lines, x = 0 and y = 0, are the asymptotes, and the gap at x = 0 is why this is the one toolkit function whose graph comes in two disconnected pieces.

Section 5Transformations

Moving a graph without redrawing it

A transformation changes a graph's position or proportions while keeping its essential shape. Once you know the six toolkit shapes, transformations let you graph hundreds of functions without ever plotting a point. The whole system rests on three moves: shifts (slide it), reflections (flip it), and stretches (pull or squash it).

There is one quirk that confuses everyone the first time, so let us name it up front. Changes made outside the function do what you expect — add 3 outside and the graph goes up 3. Changes made inside the parentheses, next to the x, do the opposite of what you expect — you see x − 3 inside and the graph goes right 3, not left.

Why? Because the input change happens before the rule runs. In f(x − 3), to get the output that f used to give at 0, you now need x − 3 = 0, meaning x = 3. The interesting behavior moved to a larger x. It moved right. Understanding that one sentence beats memorizing the rule.

Worked example 11 — decoding a full transformation

Describe the graph of g(x) = −2(x − 3)² + 5 as a sequence of transformations of f(x) = x², then verify with points.

Step 1. Identify the parent. The squared quantity tells you this starts as the parabola f(x) = x².

Step 2. Read the inside first: (x − 3)². Inside means horizontal, and the sign flips, so this is a shift right 3. The vertex leaves (0, 0) and lands at (3, 0).

Step 3. Read the multiplier: the 2 in −2. That is outside, so it is a vertical stretch by a factor of 2 — every height doubles, making the parabola narrower.

Step 4. Read the negative sign in −2. That is a reflection across the x-axis, so the parabola now opens downward.

Step 5. Read the +5 on the outside: a shift up 5. The vertex moves from (3, 0) to (3, 5).

Step 6. Verify by evaluating. This is the step that makes the answer trustworthy rather than hopeful.
g(3) = −2(3 − 3)² + 5 = −2(0)² + 5 = −2(0) + 5 = 0 + 5 = 5. Vertex confirmed at (3, 5). ✓
g(4) = −2(4 − 3)² + 5 = −2(1)² + 5 = −2(1) + 5 = −2 + 5 = 3. Point (4, 3).
g(2) = −2(2 − 3)² + 5 = −2(−1)² + 5 = −2(1) + 5 = −2 + 5 = 3. Point (2, 3) — the mirror image of (4, 3), exactly as symmetry about x = 3 requires. ✓
g(5) = −2(5 − 3)² + 5 = −2(2)² + 5 = −2(4) + 5 = −8 + 5 = −3. Point (5, −3).
Outputs falling away on both sides of the vertex confirm the downward opening. The range is (−∞, 5], since 5 is the maximum and it is attained.

Worked example 12 — tracking one point through

The point (2, 4) lies on f(x) = x², since 2² = 4. Where does it end up on g(x) = −2(x − 3)² + 5?

Step 1. Handle the x-coordinate using the inside change. We need the inside quantity to equal the old input: x − 3 = 2, so x = 2 + 3 = 5. The new x-coordinate is 5.
Step 2. Handle the y-coordinate using the outside changes, in the order they act. Start with the old output 4. Multiply by −2: 4(−2) = −8. Then add 5: −8 + 5 = −3.
Step 3. The image point is (5, −3).
Step 4. Check against the direct computation from example 11: we already found g(5) = −3. The two methods agree. ✓

One more, with a different parent. For h(x) = |x + 4| − 2: the inside x + 4 shifts left 4, and the outside −2 shifts down 2, so the V's corner moves from (0, 0) to (−4, −2). Check: h(−4) = |−4 + 4| − 2 = |0| − 2 = 0 − 2 = −2 ✓. And h(0) = |0 + 4| − 2 = |4| − 2 = 4 − 2 = 2, so (0, 2) is on the graph — four units right of the corner and four units up, exactly the slope-1 arm of a V.

ReferenceWords worth knowing

A quick reference for the vocabulary in this chapter — skim it now, and circle back any time a word slows you down.

Absolute value function

The toolkit function f(x) = |x|, whose graph is a V with its corner at the origin; domain (−∞, ∞), range [0, ∞).

Asymptote

An invisible line a graph gets arbitrarily close to as x or y grows without bound, such as x = 0 and y = 0 for f(x) = 1/x. The graph of 1/x never touches either one, but in general a graph may cross a horizontal asymptote before settling onto it.

Cubic function

The toolkit function f(x) = x³, an S-shaped curve through the origin; domain and range are both all real numbers.

Dependent variable

The output variable, usually y or f(x), whose value is determined by the input.

Domain

The set of all inputs a function is allowed to accept.

Function

A relation that assigns exactly one output to each input.

Function notation

Writing a rule as f(x), read "f of x," where f names the function and the parentheses hold the input.

Horizontal shift

A sideways slide produced by a change inside the function, f(x − h); the graph moves opposite the sign, so x − 3 moves it right 3.

Independent variable

The input variable, usually x, whose value you are free to choose.

Interval notation

Shorthand for a set of numbers using brackets to include endpoints and parentheses to exclude them, as in [4, ∞).

Linear function

A function whose graph is a straight line; the toolkit version is the identity function f(x) = x.

Mapping diagram

A picture with arrows from each input to its output; it shows a function when exactly one arrow leaves each input.

Quadratic function

The toolkit function f(x) = x² and its transformations; the graph is a parabola.

Range

The set of all outputs a function actually produces.

Reciprocal function

The toolkit function f(x) = 1/x, graphed as two branches with asymptotes at x = 0 and y = 0.

Reflection

A flip of the graph; −f(x) flips it across the x-axis and f(−x) flips it across the y-axis.

Relation

Any pairing of inputs with outputs; every function is a relation, but not every relation is a function.

Square root function

The toolkit function f(x) = √x; domain [0, ∞), range [0, ∞).

Toolkit functions

The six standard parent graphs — x, x², |x|, √x, x³, and 1/x — that later functions are built from.

Transformation

A change of a graph's position or proportions — a shift, reflection, stretch, or compression — that preserves its basic shape.

Vertical line test

A graph represents a function if and only if no vertical line intersects it more than once.

Vertical stretch

Multiplying outputs by a > 1 in a·f(x), making the graph taller and narrower; 0 < a < 1 compresses it instead.

Systems of Equations & Inequalities

Up to now, almost every equation you have solved had one unknown hiding in it. Real questions are rarely that tidy: two prices, two speeds, two concentrations, two unknown quantities tangled together in the same story. A system of equations is simply two or more equations that have to be true at the same time, and solving one means finding the values that satisfy all of them at once. This chapter gives you four reliable ways to do that — graphing, substitution, elimination, and a systematic method for three unknowns — and then puts them to work on mixture problems, rate problems, and shaded regions on a graph. Every method here is arithmetic you already know, arranged in a careful order; go slowly, write every line down, and it will hold.

Section 1Solving by graphing

What a "solution" means when there are two equations

A system of equations is a set of equations that must all be true at the same moment. When we write

x + y = 6
y = 2x − 3

we are asking a single question: is there a pair of numbers (x, y) that makes both sentences true? A solution of a system in two variables is an ordered pair, not a single number. The pair (3, 3) is a solution only if it works in the first equation and in the second. Checking one and stopping is the single most common way students lose points on this topic, so build the habit now: always check in every original equation.

Here is the geometric idea that makes graphing work. The graph of a linear equation is the picture of all the pairs that satisfy it. So the graph of equation one shows every pair that works in equation one; the graph of equation two shows every pair that works in equation two. A pair that works in both must lie on both graphs — it must be a point where the two lines intersect. That is the whole method.

A fully worked example

Solve by graphing:

Equation 1: x + y = 6
Equation 2: 2x − y = 3

Step 1 — put each line in slope-intercept form, y = mx + b, because that form is the easiest to graph.

Equation 1: x + y = 6. Subtract x from both sides: y = −x + 6. Slope m = −1, y-intercept b = 6.

Equation 2: 2x − y = 3. Subtract 2x from both sides: −y = −2x + 3. Multiply both sides by −1: y = 2x − 3. Slope m = 2, y-intercept b = −3.

Step 2 — build a small table of points for each line. Pick easy x-values and do the arithmetic out loud.

Step 3 — read the intersection. The point (3, 3) shows up on both lines, so that is where the graphs cross.

Step 4 — check in both original equations. Equation 1: 3 + 3 = 6, and 6 = 6 ✓. Equation 2: 2(3) − 3 = 6 − 3 = 3, and 3 = 3 ✓. Both are true, so the solution is (3, 3).

Graphing is honest but imprecise: if the true intersection were at (2.37, 1.14) you would never read it off graph paper. That is exactly why the algebraic methods in Sections 2 and 3 exist. Use graphing to understand what you are doing and to sanity-check an answer; use algebra to get the exact numbers.

Three pictures, three outcomes

Two lines in a plane can only relate to each other in three ways, so a two-equation linear system has only three possible outcomes. Learning to spot which one you have — before you do any work — saves a lot of time.

Case A: they cross once. Different slopes force the lines to meet at exactly one point. Example: y = −x + 6 and y = 2x − 3 have slopes −1 and 2. One solution. Such a system is called consistent (it has at least one solution) and independent (the equations say genuinely different things).

Case B: they are parallel. Consider y = 3x + 1 and y = 3x − 4. Both have slope 3, but their y-intercepts are 1 and −4, so they are different lines running side by side forever. They never meet, so no ordered pair satisfies both. The system is inconsistent; the solution set is empty. You can see the contradiction algebraically too: if y equals both, then 3x + 1 = 3x − 4, and subtracting 3x from both sides gives 1 = −4, which is false. No x can fix that.

Case C: they are the same line. Consider 2x + 4y = 8 and x + 2y = 4. Divide every term of the first by 2: (2x)/(2) + (4y)/(2) = (8)/(2), which is x + 2y = 4 — the second equation exactly. These are two disguises for one line, so they are coincident. Every point on that line satisfies both equations, and there are infinitely many solutions. Such a system is consistent and dependent. In slope-intercept form both become y = −(1)/(2)x + 2: same slope, same intercept.

Section 2Substitution

The idea: trade one variable for an equal expression

The trouble with a system is that each equation has two unknowns, and you know how to solve equations with only one unknown. The substitution method removes that trouble directly: solve one equation for one variable, then replace that variable in the other equation with the expression it equals. Because you are swapping in something genuinely equal, the new equation is still true — but now it has only one variable, and you are back on familiar ground.

The procedure is always the same four moves:

A warning on step 2: you must substitute into the other equation. If you substitute back into the same equation you came from, everything cancels and you get a true-but-useless statement like 6 = 6. That is not an error in the math; it just means you told the equation something it already knew.

Worked example 1 — a variable is already isolated

Solve:   y = 3x − 4   and   2x + y = 11.

Step 1. The first equation is already solved for y: y = 3x − 4. Nothing to do.

Step 2. Substitute (3x − 4) in place of y in the second equation:

2x + y = 11  →  2x + (3x − 4) = 11

Step 3. Solve. Combine like terms: 2x + 3x = 5x, so the equation is 5x − 4 = 11. Add 4 to both sides: 5x = 15. Divide both sides by 5: x = 3.

Step 4. Back-substitute x = 3 into the isolated equation: y = 3(3) − 4 = 9 − 4 = 5.

Check both originals. First: y = 3x − 4 becomes 5 = 3(3) − 4 = 5 ✓. Second: 2x + y = 2(3) + 5 = 6 + 5 = 11 ✓. The solution is (3, 5).

Worked example 2 — when you have to isolate first

Solve:   x − 2y = 1   and   3x + 4y = 23.

Step 1 — choose wisely. Look for a variable whose coefficient is 1 or −1, because isolating it will not create fractions. In the first equation x has coefficient 1. Add 2y to both sides of x − 2y = 1:  x = 2y + 1.

Step 2 — substitute into the other equation. Replace x with (2y + 1):

3x + 4y = 23  →  3(2y + 1) + 4y = 23

Step 3 — solve. Distribute the 3: 3(2y) = 6y and 3(1) = 3, so we have 6y + 3 + 4y = 23. Combine like terms: 6y + 4y = 10y, giving 10y + 3 = 23. Subtract 3 from both sides: 10y = 20. Divide by 10: y = 2.

Step 4 — back-substitute. Using x = 2y + 1: x = 2(2) + 1 = 4 + 1 = 5.

Check both originals. First: x − 2y = 5 − 2(2) = 5 − 4 = 1 ✓. Second: 3x + 4y = 3(5) + 4(2) = 15 + 8 = 23 ✓. The solution is (5, 2).

When the variables all disappear

Sometimes every variable cancels partway through. Do not panic — the sentence you are left with is telling you which of the three pictures from Section 1 you are in.

All variables gone, statement false. Solve y = 2x + 1 and 4x − 2y = 5. Substitute: 4x − 2(2x + 1) = 5. Distribute the −2: −2(2x) = −4x and −2(1) = −2, so 4x − 4x − 2 = 5, which simplifies to −2 = 5. That is false for every value of x, so there is no solution — the lines are parallel.

All variables gone, statement true. Solve y = 2x + 1 and 6x − 3y = −3. Substitute: 6x − 3(2x + 1) = −3, so 6x − 6x − 3 = −3, which simplifies to −3 = −3. That is true for every value of x, so there are infinitely many solutions — the two equations describe the same line. We report the answer as "all ordered pairs satisfying y = 2x + 1."

Substitution is not only for lines

Substitution is the method that keeps working when an equation is not linear, which is why it is worth learning well. Solve y = x² − 1 and y = x + 1. Since both expressions equal y, they equal each other:

x² − 1 = x + 1

Subtract x and subtract 1 from both sides: x² − x − 2 = 0. Factor: (x − 2)(x + 1) = 0. So x = 2 or x = −1. Back-substitute into y = x + 1: when x = 2, y = 3; when x = −1, y = 0. Check in the parabola: 2² − 1 = 3 ✓ and (−1)² − 1 = 1 − 1 = 0 ✓. The line crosses the parabola twice, at (2, 3) and (−1, 0).

Two cautions come with nonlinear systems. First, a variable may carry a restriction, and you should write it down before you solve, not after. If an equation contains (1)/(x − 3), the denominator cannot be zero, so x ≠ 3 is an excluded value and a "solution" of x = 3 must be thrown out no matter how cleanly the algebra produced it. Here is that trap in miniature: solve y = (x)/(x − 2) and y = (2)/(x − 2). Both denominators force x ≠ 2. Setting the two expressions equal and multiplying both sides by (x − 2) gives x = 2 — the one value the system forbids. That candidate is extraneous, and the honest answer is no solution.

Second, if solving ever requires squaring both sides or multiplying through by an expression containing a variable, the new equation can pick up extraneous solutions that do not satisfy the original, because neither move is reversible. The same caution applies to any logarithmic equation you meet in a system: a logarithm is defined only for a positive argument, so log expressions carry a built-in domain restriction and every candidate must be tested against it. None of this is a reason to fear the method — it is a reason to check every candidate in the original equations, which you were going to do anyway.

Section 3Elimination

The idea: add the equations so a variable cancels

The elimination method — also called the addition method — rests on a fact that feels almost too easy: if A = B and C = D, then A + C = B + D. You may add two true equations together, left side to left side and right side to right side, and the result is still true. If the coefficients of one variable are opposites, that variable adds to zero and vanishes, leaving one equation in one unknown.

When the coefficients are not already opposites, you make them opposites. Multiplying an entire equation by a nonzero number does not change its solution set — 2x + 3y = 7 and 4x + 6y = 14 have exactly the same solutions — so you are free to scale either equation, or both, until the numbers line up.

Worked example 1 — the coefficients already cancel

Solve:   2x + 5y = 23   and   3x − 5y = −3.

Step 1 — look for opposites. The y-terms are +5y and −5y. Adding them gives 5y + (−5y) = 0y = 0. The y disappears with no extra work.

Step 2 — add the equations column by column.

The arithmetic on the constants: 23 + (−3) = 20. On the x-terms: 2x + 3x = 5x. So the sum is 5x = 20.

Step 3 — solve. Divide both sides by 5: x = 4.

Step 4 — back-substitute. Put x = 4 into equation 1: 2(4) + 5y = 23, so 8 + 5y = 23. Subtract 8 from both sides: 5y = 15. Divide by 5: y = 3.

Check both originals. Equation 1: 2(4) + 5(3) = 8 + 15 = 23 ✓. Equation 2: 3(4) − 5(3) = 12 − 15 = −3 ✓. The solution is (4, 3).

Worked example 2 — multiply first, then eliminate

Solve:   3x + 4y = 10   and   2x + 5y = 9.

Step 1 — decide what to eliminate. Nothing cancels yet. Target x: the coefficients are 3 and 2, and the least common multiple of 3 and 2 is 6. If we can make them +6x and −6x, they will cancel.

Step 2 — multiply each equation through. Multiply every term of equation 1 by 2:

2(3x) + 2(4y) = 2(10)  →  6x + 8y = 20

Multiply every term of equation 2 by −3:

−3(2x) + (−3)(5y) = −3(9)  →  −6x − 15y = −27

Step 3 — add. The x-terms: 6x + (−6x) = 0. The y-terms: 8y + (−15y) = −7y. The constants: 20 + (−27) = −7. So −7y = −7.

Step 4 — solve. Divide both sides by −7: y = 1.

Step 5 — back-substitute into an original equation. Use 3x + 4y = 10 with y = 1: 3x + 4(1) = 10, so 3x + 4 = 10. Subtract 4: 3x = 6. Divide by 3: x = 2.

Check both originals. Equation 1: 3(2) + 4(1) = 6 + 4 = 10 ✓. Equation 2: 2(2) + 5(1) = 4 + 5 = 9 ✓. The solution is (2, 1).

Notice that back-substitution used an original equation, not one of the multiplied versions. Either would work, but the originals are the ones you must satisfy, so using them doubles as part of your check.

The same two special endings

Elimination announces parallel and coincident lines the same way substitution does. Solve 2x + 3y = 7 and 4x + 6y = 5. Multiply the first equation by −2: −4x − 6y = −14. Add it to the second: (−4x + 4x) + (−6y + 6y) = −14 + 5, which is 0 = −9. False, so there is no solution.

Now try 2x + 3y = 7 and 4x + 6y = 14. Multiply the first by −2 again: −4x − 6y = −14. Add: 0 = −14 + 14 = 0. True, so there are infinitely many solutions — the second equation is just the first one doubled.

Which method should you reach for?

Both methods always work. The question is only which one costs less arithmetic, and the answer is usually visible in the coefficients.

Section 4Systems in three variables

What changes and what does not

An equation like 2x − y + 3z = 9 is linear in three variables. Its solution is no longer an ordered pair but an ordered triple (x, y, z), and its graph is no longer a line but a flat plane floating in three-dimensional space. A system of three such equations asks: is there a point that lies on all three planes at once?

Three planes can meet at exactly one point (like the corner of a room where two walls and the floor come together) — that is the unique solution case. They can share a whole line or a whole plane, giving infinitely many solutions. Or they can have no common point at all — two of them parallel, or three of them arranged like the sides of a triangular prism — giving no solution.

The strategy is comforting because it reuses what you already know: knock the 3 × 3 system down to a 2 × 2 system, solve that, then climb back up. The tool for knocking it down is elimination.

A fully worked 3 × 3 example

Solve:

E1:  x + y + z = 6
E2:  2x − y + 3z = 9
E3:  x − y + z = 2

Step 1 — choose a variable to eliminate everywhere. Scan the y-column: +1y, −1y, −1y. Those are already opposites, so y is the cheapest target. We will eliminate y twice, using two different pairs of equations.

Step 2 — eliminate y from E1 and E2 by adding them.

(x + 2x) + (y − y) + (z + 3z) = 6 + 9

The x-terms give 3x; the y-terms give 0; the z-terms give 4z; the constants give 15. Call the result Equation A:  3x + 4z = 15.

Step 3 — eliminate y from E1 and E3 by adding them.

(x + x) + (y − y) + (z + z) = 6 + 2

That gives 2x + 2z = 8. Divide every term by 2 to keep the numbers small: Equation B:  x + z = 4.

Equations A and B form an ordinary 2 × 2 system in x and z — the hard part is over.

Step 4 — solve the 2 × 2 system. Equation B has a coefficient of 1, so substitution is cheap. From B: x = 4 − z. Substitute into A:

3(4 − z) + 4z = 15

Distribute the 3: 3(4) = 12 and 3(−z) = −3z, so 12 − 3z + 4z = 15. Combine the z-terms: −3z + 4z = 1z, giving 12 + z = 15. Subtract 12 from both sides: z = 3.

Step 5 — back-substitute for the second variable. Using x = 4 − z:  x = 4 − 3 = 1.

Step 6 — back-substitute into an original three-variable equation for the last one. Use E1: x + y + z = 6 becomes 1 + y + 3 = 6, so 4 + y = 6. Subtract 4: y = 2.

Step 7 — check in all three originals. This is not optional in a 3 × 3, because a single arithmetic slip in step 2 quietly poisons everything after it.

All three hold, so the solution is the ordered triple (1, 2, 3).

Bookkeeping rules that keep you out of trouble

Three-variable systems are not conceptually harder than two-variable ones; they are just longer, and length is where errors live. A few habits carry most of the weight.

Eliminate the same variable both times. If you cancel y from the pair (E1, E2) and then cancel z from the pair (E1, E3), your two new equations will not share a common pair of variables and you will have gained nothing.

Use each equation at least once. A classic trap is to use E1 with E2, then E1 with E2 again in a different disguise. You must bring E3 into the work, or its information never enters the answer.

Aim for triangular form if you like structure. Some people prefer to rearrange the system into a staircase — the first equation with all three variables, the second with two, the third with one — and then solve from the bottom up. That process is called back-substitution, and it is exactly what steps 4 through 6 above did, just written differently.

Read the endings the same way as before. If eliminating produces a false statement such as 0 = 7, the system is inconsistent and has no solution. If it produces 0 = 0, the equations are dependent and there are infinitely many solutions — the planes share a line or a plane rather than a point.

Section 5Applications and systems of inequalities

Turning a story into two equations

Word problems become manageable the moment you accept a rigid routine. Name the unknowns with letters and units. Write one equation for each independent fact in the story. Two unknowns need two facts; three unknowns need three. Then solve by whichever method the coefficients invite, and — the step people skip — ask whether the answer makes sense in the story. A negative number of liters or a fractional number of chairs is a signal to go back and reread, not a final answer.

Two families of problems show up constantly, so it is worth seeing each worked completely.

Worked example 1 — a mixture problem

A technician has a 20% saline solution and a 50% saline solution. How many milliliters of each are needed to make 300 mL of a 30% solution?

Step 1 — name the unknowns. Let x = milliliters of the 20% solution and y = milliliters of the 50% solution. Both must be at or above zero: x ≥ 0 and y ≥ 0.

Step 2 — write one equation for total volume. The two amounts have to add up to 300 mL:

x + y = 300

Step 3 — write one equation for the amount of salt. This is the equation people forget. Track the substance, not the liquid. The 20% bottle contributes 0.20x mL of salt, the 50% bottle contributes 0.50y mL, and the finished 300 mL at 30% contains 0.30(300) = 90 mL of salt.

0.20x + 0.50y = 90

Step 4 — clear the decimals. Multiply every term of the salt equation by 10: 10(0.20x) = 2x, 10(0.50y) = 5y, and 10(90) = 900. So 2x + 5y = 900.

Step 5 — solve. The volume equation has coefficients of 1, so substitution is cheap. From x + y = 300 we get x = 300 − y. Substitute into 2x + 5y = 900:

2(300 − y) + 5y = 900

Distribute: 2(300) = 600 and 2(−y) = −2y, giving 600 − 2y + 5y = 900. Combine the y-terms: −2y + 5y = 3y, so 600 + 3y = 900. Subtract 600 from both sides: 3y = 300. Divide by 3: y = 100.

Step 6 — back-substitute. x = 300 − 100 = 200.

Step 7 — check against the story, not just the algebra. Volume: 200 + 100 = 300 mL ✓. Salt: 0.20(200) + 0.50(100) = 40 + 50 = 90 mL ✓. Concentration of the result: (90)/(300) = 0.30 = 30% ✓. Both amounts are positive, so they are physically possible. Answer: 200 mL of the 20% solution and 100 mL of the 50% solution.

Sanity check on the reasoning: 30% sits closer to 20% than to 50%, so we should need more of the weak solution than the strong one — and we do, 200 mL versus 100 mL. That kind of glance catches sign errors before they reach your paper.

Worked example 2 — a rate problem

A boat travels 36 miles downstream in 2 hours, then returns the same 36 miles upstream in 3 hours. Find the boat's speed in still water and the speed of the current.

Step 1 — name the unknowns. Let b = the boat's speed in still water (mph) and c = the current's speed (mph).

Step 2 — model the two directions. Going downstream the current helps, so the effective speed is b + c. Going upstream the current fights the boat, so the effective speed is b − c.

Step 3 — use distance = rate × time, rearranged as rate = distance ÷ time.

Downstream: b + c = (36)/(2) = 18
Upstream: b − c = (36)/(3) = 12

Step 4 — eliminate. The c-terms are +c and −c, already opposites. Add the two equations: (b + b) + (c − c) = 18 + 12, which is 2b = 30. Divide both sides by 2: b = 15.

Step 5 — back-substitute. Put b = 15 into b + c = 18: 15 + c = 18, so c = 3.

Step 6 — check against the story. Downstream speed 15 + 3 = 18 mph for 2 hours covers 18 × 2 = 36 miles ✓. Upstream speed 15 − 3 = 12 mph for 3 hours covers 12 × 3 = 36 miles ✓. Both speeds are positive, and the current is slower than the boat — otherwise the boat could not go upstream at all. Answer: the boat travels 15 mph in still water and the current runs at 3 mph.

Systems of inequalities: shading instead of pinpointing

Change an equals sign to an inequality sign and the answer stops being a point and becomes a region. The graph of y ≤ −x + 6 is not a line; it is the line together with every point below it. Graphing a linear inequality takes three moves.

Move 1 — graph the boundary line by pretending the inequality sign is an equals sign. Draw it solid if the symbol is ≤ or ≥, because those points are included. Draw it dashed if the symbol is < or >, because those points are excluded.

Move 2 — pick a test point that is not on the line. Use (0, 0) whenever the line misses the origin, because the arithmetic is trivial. If the line passes through the origin, use something like (1, 0) or (0, 1) instead.

Move 3 — shade. If the test point makes the inequality true, shade the side containing it. If it makes the inequality false, shade the other side.

One algebra warning first. If you have to solve for y and that requires multiplying or dividing both sides by a negative number, the inequality symbol flips. Adding or subtracting never flips it; only multiplying or dividing by a negative does. Take 2x − 3y ≥ 6. Subtract 2x: −3y ≥ −2x + 6. Now divide every term by −3, and because −3 is negative, ≥ becomes ≤:  y ≤ (2)/(3)x − 2. Multiplying works the same way: from −(1)/(2)y < 4, multiply both sides by −2 and < flips to >, giving y > −8. Forgetting that flip shades exactly the wrong half of the plane.

Worked example 3 — graphing a system of two inequalities

Graph the solution region of:   y ≤ −x + 6   and   y ≥ 2x − 3.

Step 1 — handle the first inequality. Boundary: y = −x + 6, the same line from Section 1, passing through (0, 6) and (6, 0). The symbol is ≤, so the line is solid. Test (0, 0): is 0 ≤ −(0) + 6? That asks whether 0 ≤ 6, which is true, so shade the side containing the origin — the region below the line.

Step 2 — handle the second inequality. Boundary: y = 2x − 3, passing through (0, −3) and (2, 1). The symbol is ≥, so this line is also solid. Test (0, 0): is 0 ≥ 2(0) − 3? That asks whether 0 ≥ −3, which is true, so shade the side containing the origin — the region above this line.

Step 3 — find the overlap. The solution of the system is only where both shadings agree. The two boundary lines cross at (3, 3) — we computed that in Section 1 — so the solution region is a wedge with its point at (3, 3), opening to the left, bounded above by y = −x + 6 and below by y = 2x − 3. Both edges belong to the region because both lines are solid.

Step 4 — verify with sample points. Try (0, 0): 0 ≤ 6 ✓ and 0 ≥ −3 ✓, so the origin is in the region, as the shading says. Try (5, 5): is 5 ≤ −5 + 6 = 1? No — 5 ≤ 1 is false, so (5, 5) is outside, which matches the picture. Try (3, 3), the corner: 3 ≤ −3 + 6 = 3 ✓ (equality is allowed) and 3 ≥ 2(3) − 3 = 3 ✓. The corner point is included.

A system of inequalities can also have no solution. Graph y > x + 2 and y < x − 1. Both boundaries have slope 1, so they are parallel, and the first inequality wants points above the higher line while the second wants points below the lower line. Those two regions never overlap, so the solution set is empty. You can see it algebraically as well: the two conditions together would require x + 2 < y < x − 1, and no number is both greater than x + 2 and less than x − 1.

ReferenceWords worth knowing

A quick reference for the vocabulary in this chapter — skim it now, and come back whenever a word slows you down mid-problem.

Back-substitution

Taking a value you have just found and putting it back into an earlier equation to find the remaining variable(s); the climb back up after reducing a system.

Boundary line

The line you get by replacing an inequality's symbol with an equals sign; drawn solid for ≤ or ≥ and dashed for < or >.

Break-even point

The solution of the system cost = revenue; the production level at which money in equals money out.

Coincident lines

Two equations that graph as the same line; they have identical slope and identical y-intercept and produce infinitely many solutions.

Consistent system

A system that has at least one solution.

Dependent system

A consistent system whose equations carry the same information (one is a multiple of the other), giving infinitely many solutions.

Elimination method

Solving a system by adding the equations — after multiplying one or both if necessary — so that one variable's terms cancel to zero. Also called the addition method.

Excluded value

A number the variable is not allowed to take because it would make a denominator zero (or break another domain rule, such as a logarithm needing a positive argument); it must be stated before solving and any candidate equal to it must be rejected.

Extraneous solution

A value produced by a legitimate algebraic step — squaring both sides, or multiplying through by an expression containing a variable — that does not actually satisfy the original equation; the reason every candidate is checked in the originals.

Half-plane

One of the two regions a line cuts the coordinate plane into; the solution set of a linear inequality is a half-plane, with or without its boundary.

Inconsistent system

A system with no solution at all; in two variables, parallel lines. Algebraically it announces itself as a false statement such as 0 = −9.

Independent system

A system whose equations give genuinely different information; in two variables, lines with different slopes and exactly one solution.

Linear equation in two variables

An equation that can be written Ax + By = C with A and B not both zero; its graph is a straight line.

Ordered pair

A solution written (x, y) with the order fixed; (3, 5) and (5, 3) are different points.

Ordered triple

A solution in three variables written (x, y, z); it must satisfy all three equations.

Parallel lines

Distinct lines with the same slope and different y-intercepts; they never intersect, so the system is inconsistent.

Slope-intercept form

The form y = mx + b, where m is the slope and b the y-intercept; the most convenient form for graphing and for comparing two lines.

Solution of a system

Any set of values that makes every equation in the system true simultaneously.

Substitution method

Solving a system by isolating one variable and replacing it in the other equation with the expression it equals.

System of equations

Two or more equations considered together, whose solutions must satisfy all of them at once.

System of inequalities

Two or more inequalities considered together; the solution is the overlap of their shaded regions, and may be empty.

Test point

Any point not on the boundary line, substituted into an inequality to decide which side to shade; (0, 0) is the usual choice when the line misses the origin.

Triangular form

A rearrangement of a system in which each successive equation has one fewer variable, so it can be solved from the bottom up by back-substitution.

Polynomials & Factoring

A polynomial is nothing more exotic than a string of terms added together, each one a number multiplied by a variable raised to a whole-number power. You have been handling them since your first algebra class — 2x + 5 is one, and so is x² − 3x − 4. This chapter teaches you to add them, subtract them, multiply them, and then — the part that makes people nervous — run the multiplication backward, which is called factoring. Factoring is a skill, not a talent: it is a short list of patterns you learn to recognize, tried in a fixed order, every single time. Work the examples with a pencil rather than reading them, and by the end of this chapter the list will feel automatic.

Section 1Polynomials

What counts as a polynomial

A term is a number multiplied by one or more variables raised to powers: 7x³, −2x, and 15 are all terms. A polynomial is a sum of such terms where every exponent on a variable is a whole number (0, 1, 2, 3, …) and no variable ever sits in a denominator or under a radical. That last sentence is the whole definition, and it is worth memorizing, because the entire toolkit in this chapter only works on polynomials.

So 4x³ − x + 9 is a polynomial. But (3)/(x) is not, because (3)/(x) is really 3x⁻¹ and −1 is not a whole number. Neither is √x, which is x^1/2. Neither is 2^x, where the variable is up in the exponent instead of down in the base. Those expressions are perfectly legal mathematics — they are just not polynomials, and the factoring patterns below do not apply to them.

The number multiplying a term is its coefficient. A term with no variable at all — a lone number — is the constant term. In 4x³ − x + 9, the coefficients are 4, −1, and 9, and 9 is the constant term. Notice that the coefficient of x is −1, not 1: the sign in front of a term belongs to that term. Half of all sign errors in algebra come from forgetting that one rule.

Degree, standard form, and names

The degree of a term is the exponent on its variable (or, if a term has several variables, the sum of the exponents: 5x²y³ has degree 2 + 3 = 5). The degree of the polynomial is the largest degree among its terms. A nonzero constant like 9 has degree 0, because 9 = 9x⁰.

A polynomial is in standard form when its terms are written from highest degree down to lowest. The coefficient of that first, highest-degree term is the leading coefficient. Rewriting 5 − 2x³ + x² in standard form gives −2x³ + x² + 5, so the degree is 3 and the leading coefficient is −2. Standard form is not fussiness for its own sake — it makes like terms line up, which is exactly what addition and subtraction need.

Adding and subtracting: combine like terms

Like terms have exactly the same variable raised to exactly the same power. 3x² and −8x² are like terms; 3x² and 3x³ are not. You combine like terms by adding their coefficients and keeping the variable part untouched, because that is just the distributive property in reverse: 3x² − 8x² = (3 − 8)x² = −5x².

Worked example — addition. Simplify (3x³ − 5x + 7) + (x³ + 4x² − 2x − 9).

1. Drop the parentheses. Because this is addition, nothing changes sign: 3x³ − 5x + 7 + x³ + 4x² − 2x − 9.
2. Collect the x³ terms. 3x³ + 1x³. Coefficients: 3 + 1 = 4. So 4x³.
3. Collect the x² terms. The first polynomial has none, so this is 0 + 4 = 4. So 4x².
4. Collect the x terms. −5x and −2x. Coefficients: −5 + (−2) = −7. So −7x.
5. Collect the constants. 7 + (−9) = −2.
6. Write in standard form. 4x³ + 4x² − 7x − 2.

Worked example — subtraction. Simplify (4x² − x + 6) − (2x² + 5x − 3).

1. Distribute the minus sign to every term inside the second parentheses. The minus sign is really −1 multiplying the whole group: −1(2x²) = −2x², −1(+5x) = −5x, and −1(−3) = +3.
2. Rewrite with no parentheses. 4x² − x + 6 − 2x² − 5x + 3.
3. Combine x² terms. 4 − 2 = 2, giving 2x².
4. Combine x terms. −1 − 5 = −6, giving −6x.
5. Combine constants. 6 + 3 = 9.
6. Answer. 2x² − 6x + 9.

Notice that the third term went from −3 to +3. If you had written 6 − 3 = 3 there, you would have gotten 2x² − 6x + 3, which is wrong. There is no partial credit from a graphing calculator, so build the habit now: a subtraction sign in front of a group flips every sign inside the group.

Section 2Multiplying polynomials

The one rule: everybody meets everybody

All polynomial multiplication is a single idea repeated: each term in the first factor multiplies each term in the second factor, and then you combine like terms. That is the distributive property, a(b + c) = ab + ac, applied as many times as needed. When you multiply the variable parts, remember the exponent rule x^m · xⁿ = x^m+n — here the exponents do add.

Worked example — monomial times polynomial. Multiply 3x²(4x³ − 2x + 5).

1. 3x² · 4x³: coefficients 3 · 4 = 12; variables x² · x³ = x²⁺³ = x⁵. Result: 12x⁵.
2. 3x² · (−2x): coefficients 3 · (−2) = −6; variables x² · x¹ = x³. Result: −6x³.
3. 3x² · 5 = 15x².
4. Answer. 12x⁵ − 6x³ + 15x². No like terms to combine, since the degrees 5, 3, and 2 are all different.

FOIL: a nickname for one particular case

When both factors are binomials, "everybody meets everybody" produces exactly four products, and the mnemonic FOIL keeps you from missing one: First terms, Outer terms, Inner terms, Last terms. Understand that FOIL is not a separate rule — it is the distributive property with a name, and it only applies to two-term times two-term.

Worked example — FOIL. Multiply (2x + 3)(x − 5).

1. First: 2x · x = 2x².
2. Outer: 2x · (−5) = −10x.
3. Inner: 3 · x = 3x.
4. Last: 3 · (−5) = −15.
5. String them together. 2x² − 10x + 3x − 15.
6. Combine the middle terms. −10 + 3 = −7, so the answer is 2x² − 7x − 15.

Worked example — binomial times trinomial. Multiply (x + 2)(x² − 3x + 4). FOIL does not apply here — there are six products, not four — so distribute deliberately.

1. Send x through the trinomial. x · x² = x³; x · (−3x) = −3x²; x · 4 = 4x.
2. Send 2 through the trinomial. 2 · x² = 2x²; 2 · (−3x) = −6x; 2 · 4 = 8.
3. Line up all six products. x³ − 3x² + 4x + 2x² − 6x + 8.
4. x² terms: −3 + 2 = −1, giving −x². x terms: 4 − 6 = −2, giving −2x.
5. Answer. x³ − x² − 2x + 8.

Three special products worth memorizing

Three particular products come up so often — and are so useful read backward as factoring patterns in Section 5 — that you should know them cold rather than FOIL them out every time.

Worked example — verifying the first pattern. Expand (3x + 5)² the long way to see where 2ab comes from. Write it as (3x + 5)(3x + 5) and FOIL: First, 3x · 3x = 9x². Outer, 3x · 5 = 15x. Inner, 5 · 3x = 15x. Last, 5 · 5 = 25. Adding, 9x² + 15x + 15x + 25 = 9x² + 30x + 25. The Outer and Inner products are identical, which is precisely why the middle term is twice the product of the two pieces.

Worked example — why the middle vanishes. Expand (4x − 7)(4x + 7). First, 4x · 4x = 16x². Outer, 4x · 7 = 28x. Inner, −7 · 4x = −28x. Last, −7 · 7 = −49. The middle terms are opposites: 28x − 28x = 0. What survives is 16x² − 49.

Section 3Factoring out the GCF and by grouping

Factoring is multiplication run backward

To factor a polynomial is to write it as a product. You already know how to check your work perfectly: multiply the factors back out and see whether you land on what you started with. That check costs fifteen seconds and eliminates essentially all factoring errors, so do it every time.

Factoring matters because a product equal to zero gives you information a sum never can. If AB = 0, then A = 0 or B = 0 — the zero-product property — and that is how factored form turns into solutions in the next chapter. Factoring also lets you simplify fractions. One caution while you are here: when you cancel a common factor out of a rational expression, the excluded values do not disappear. For instance (x² − 9)/(x − 3) = ((x − 3)(x + 3))/(x − 3) = x + 3, but only for x ≠ 3, because the original expression is undefined there.

Step one, every time: the greatest common factor

The greatest common factor (GCF) of a polynomial is the largest monomial that divides evenly into every term. Build it in two pieces: the greatest common factor of the coefficients, and each shared variable raised to the lowest power that appears anywhere in the polynomial. Pulling out the GCF is always the first move — before grouping, before trinomials, before anything — because it makes every later step smaller.

Worked example. Factor 12x⁴ − 18x³ + 30x².

1. GCF of the coefficients. Factors of 12: 1, 2, 3, 4, 6, 12. Factors of 18: 1, 2, 3, 6, 9, 18. Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30. The largest number in all three lists is 6.
2. GCF of the variables. The powers present are x⁴, x³, and x². The lowest is x².
3. So the GCF is 6x².
4. Divide each term by 6x². (12x⁴)/(6x²) = 2x². (−18x³)/(6x²) = −3x. (30x²)/(6x²) = 5.
5. Write the answer. 6x²(2x² − 3x + 5).
6. Check by distributing. 6x² · 2x² = 12x⁴; 6x² · (−3x) = −18x³; 6x² · 5 = 30x². Matches the original.

When the leading coefficient is negative, it is usually worth pulling the negative out too. For −4x³ + 8x² − 12x, take the GCF as −4x: (−4x³)/(−4x) = x², (8x²)/(−4x) = −2x, and (−12x)/(−4x) = 3, so the factorization is −4x(x² − 2x + 3). Every sign inside flipped, which is exactly what dividing by a negative does.

Four terms? Try grouping

When a polynomial has four terms and no GCF shared by all four, split it into two pairs, factor the GCF out of each pair separately, and hope the leftover parentheses match. If they do, that matching parenthesis is itself a common factor and you pull it out.

Worked example. Factor x³ + 5x² + 3x + 15.

1. Group into pairs. (x³ + 5x²) + (3x + 15).
2. GCF of the first pair. x³ and 5x² share x². Factoring gives x²(x + 5).
3. GCF of the second pair. 3x and 15 share 3. Factoring gives 3(x + 5).
4. Compare the parentheses. Both are (x + 5) — the grouping worked. The expression is now x²(x + 5) + 3(x + 5).
5. Pull out the common binomial. Treat (x + 5) as a single object: it appears once in each term, so factor it out and what remains is x² + 3. Answer: (x + 5)(x² + 3).
6. Check. (x + 5)(x² + 3) = x³ + 3x + 5x² + 15 = x³ + 5x² + 3x + 15. Correct.