Chapter 3 — Front Matter

Purpose and Overview

Every major idea in Algebra I eventually depends on your ability to solve an equation. Factoring produces equations. Functions produce equations. Graphing produces equations. Word problems produce equations. The skills built in this chapter are not optional tools for one unit — they are the central engine of the entire course.

This chapter teaches equations as balanced mathematical statements. An equation does not simply ask you to calculate. It makes a claim: two expressions are equal. Your job as a student is to determine whether that claim is true, and if it involves an unknown, to find the value that makes it true.

You will begin with the simplest equations — those requiring only one operation to solve — and move progressively toward equations involving multiple steps, parentheses, variables on both sides, fractions, decimals, and real-world situations. By the end of this chapter, you will not only know how to solve equations — you will understand why each step is valid, what it means for an answer to be correct, and how to check your work in every case.

Opening Question

A shipping company charges a flat fee of $8.00 plus $3.50 per pound to deliver a package. You have a budget of exactly $29.00 for shipping. How heavy can your package be? The structure of this problem — a fixed starting cost, a rate per unit, and a known total — is an algebraic pattern that appears in finance, science, engineering, and everyday life. This chapter will show you how to write and solve that equation precisely.

14Sections

11Objectives

12Key Terms

20+Worked Examples

Learning Objectives

1Identify what an equation says and what it means for a value to be a solution.
2Check whether a given value is a solution to an equation.
3Apply the concept of balance and inverse operations to solve equations.
4Solve one-step, two-step, and multi-step linear equations in one variable.
5Solve equations that include parentheses using the Distributive Property.
6Solve equations that have variables on both sides.
7Solve equations involving fractions by clearing with the LCD.
8Solve equations involving decimals directly or by clearing with a power of 10.
9Recognize identity equations and no-solution equations.
10Write equations from verbal descriptions and solve word problems using the Read → Represent → Solve → Check → Explain method.
11Identify and correct common equation-solving mistakes.

Chapter 3 — Contents

Sections at a Glance

3.1What an Equation SaysThe equals sign as a statement of equality; structure of a linear equation; reading equations in plain language
3.2Solutions and CheckingDefinition of a solution; the checking habit; verifying by substitution into the original equation
3.3Balance and Inverse OperationsThe balance model; properties of equality; inverse operation pairs; the goal of isolation
3.4One-Step EquationsSolving by adding, subtracting, multiplying, dividing; negative coefficients; four types with examples
3.5Two-Step EquationsUndoing addition/subtraction first, then multiplication/division; negative coefficients; division in original equations
3.6Multi-Step EquationsCombining like terms first; multiple constants; the general four-step strategy
3.7Equations with ParenthesesThe Distributive Property in equations; parentheses on both sides; subtracting parenthetical expressions
3.8Equations with Variables on Both SidesCollecting variable terms on one side; choosing which side; combining distribution and collection
3.9Equations with FractionsClearing fractions with the LCD; multiplying every term; fractions with constants in numerators
3.10Equations with DecimalsSolving directly; clearing with powers of 10; tenths and hundredths; connection to finance
3.11Identity Equations and No-Solution EquationsWhat happens when variables eliminate; true constant statements; false constant statements; solution sets
3.12Writing Equations from WordsTranslation vocabulary; signal words; "less than" reversal; writing equations before solving
3.13Equation Word ProblemsRead → Represent → Solve → Check → Explain; shipping, consecutive integers, perimeter, rate and time
3.14Common Equation MistakesSeven recurring errors; explanations and corrections; building the checking habit

Section 3.1

What an Equation Says

An equation is not a question. It is a statement. The equation 3x + 5 = 17 does not ask, "What is 3x + 5?" It claims: "There exists a value of x such that 3x + 5 equals 17."

The equals sign is the heart of every equation. It is not a prompt to calculate. It is a declaration that the expression on the left and the expression on the right have the same value. In an equation, both sides are simultaneously equal. Your job is to find which value of the variable, if any, makes that simultaneous equality true.

The Structure of a Linear Equation

A linear equation in one variable is an equation that can be written in the form ax + b = c, where a, b, and c are numbers and a ≠ 0. The word "linear" means the variable appears only to the first power — no x², no √x, no 1/x. This chapter focuses exclusively on linear equations in one variable.

Equation	Plain-Language Reading
`x + 4 = 11`	Some number plus four equals eleven.
`3x = 18`	Three times some number equals eighteen.
`2x − 5 = 9`	Twice some number, minus five, equals nine.
`x/3 = 7`	Some number divided by three equals seven.

Linear Equations	Not Linear in One Variable	Reason
`x + 7 = 12`	`x² = 9`	Variable is squared (quadratic)
`5(x + 3) = 25`	`√x = 3`	Variable is under a radical
`x/4 = 9`	`xy = 10`	Two variables present

Section 3.2

Solutions and Checking

A solution to an equation is a value of the variable that makes the equation a true statement. For x + 4 = 11, the solution is x = 7 because substituting 7 gives 7 + 4 = 11, which is true. Any other value produces a false statement.

The habit of checking solutions is one of the most important skills in algebra. To check a solution: substitute the value into the original equation. Simplify both sides independently. If both sides equal the same number, the solution is correct.

Example 3.2.1

Checking a proposed solution

ProblemIs x = 4 a solution to 3x − 5 = 7?
ThinkSubstitute 4 in place of x and simplify both sides independently.
Solve3(4) − 5 = 7 12 − 5 = 7 7 = 7 ✓
CheckBoth sides equal 7.

Answer: Yes. x = 4 is a solution.

Example 3.2.2

Checking a value that is not a solution

ProblemIs x = 6 a solution to 2x + 3 = 14?
Solve2(6) + 3 = 14 12 + 3 = 14 15 ≠ 14

Answer: No. x = 6 is not a solution.

Student Warning

Always check with the original equation, not a simplified version you produced during solving. If you made an algebra error earlier, checking with a simplified form will hide the mistake.

Section 3.3

Balance and Inverse Operations

The most useful way to picture an equation is as a balance scale. The equals sign is the balance point. Whatever is on the left side has the same total weight as whatever is on the right. The key principle follows directly: if you do the same thing to both sides of a balanced scale, it stays balanced.

3x + 5

=

17

Properties of Equality

Addition: If a = b, then a + c = b + c for any number c.

Subtraction: If a = b, then a − c = b − c for any number c.

Multiplication: If a = b, then ac = bc for any number c.

Division: If a = b and c ≠ 0, then a/c = b/c.

Operation	Inverse Operation	Example
Addition	Subtraction	`x + 5 = 12 → subtract 5`
Subtraction	Addition	`x − 3 = 9 → add 3`
Multiplication	Division	`3x = 18 → divide by 3`
Division	Multiplication	`x/4 = 6 → multiply by 4`

Key Idea

You are not "moving things to the other side." You are performing operations on both sides simultaneously. The phrase "moving to the other side" is informal shorthand that causes errors when equations grow more complex. Think: balance and inverse operations.

Section 3.4

One-Step Equations

A one-step equation requires a single inverse operation. There are four types, one for each basic arithmetic operation.

Example 3.4.1

Solving by subtracting

ProblemSolve x + 9 = 15.
ThinkThe variable has 9 added to it. Undo by subtracting 9 from both sides.
Solvex + 9 = 15 x + 9 − 9 = 15 − 9 x = 6
Check6 + 9 = 15 ✓

Answer: x = 6

Example 3.4.2

Solving by adding

ProblemSolve x − 7 = 4.
ThinkThe variable has 7 subtracted from it. Undo by adding 7 to both sides.
Solvex − 7 = 4 x − 7 + 7 = 4 + 7 x = 11
Check11 − 7 = 4 ✓

Answer: x = 11

Example 3.4.3

Solving by dividing

ProblemSolve 5x = 35.
ThinkThe variable is multiplied by 5. Undo by dividing both sides by 5.
Solve5x = 35 5x/5 = 35/5 x = 7
Check5(7) = 35 ✓

Answer: x = 7

Example 3.4.4

Solving by multiplying

ProblemSolve x/4 = 9.
ThinkThe variable is divided by 4. Undo by multiplying both sides by 4.
Solvex/4 = 9 4 · (x/4) = 4 · 9 x = 36
Check36/4 = 9 ✓

Answer: x = 36

Example 3.4.5

Negative coefficient

ProblemSolve −6x = 42.
ThinkThe variable is multiplied by −6. Divide both sides by −6.
Solve−6x = 42 −6x/(−6) = 42/(−6) x = −7
Check−6(−7) = 42 ✓

Answer: x = −7

Practice 3.4One-Step Equations — Skill Practice

Solve each equation. Show your work and check your answer.

x + 14 = 23
x − 9 = 5
7x = 63
x/8 = 6
−5x = 40
x + (−3) = 12
15 = x − 4
x/(−3) = 9

Section 3.5

Two-Step Equations

A two-step equation requires two inverse operations. The standard approach: undo addition or subtraction first, then undo multiplication or division. This is the reverse of the order in which the expression was built.

Example 3.5.1

Standard two-step

ProblemSolve 2x + 5 = 13.
ThinkThe variable term is 2x. It has 5 added. Undo the addition first, then the multiplication.
Solve2x + 5 = 13 2x + 5 − 5 = 13 − 5 2x = 8 2x/2 = 8/2 x = 4
Check2(4) + 5 = 8 + 5 = 13 ✓

Answer: x = 4

Example 3.5.2

Subtraction in the original

ProblemSolve 3x − 7 = 11.
Solve3x − 7 = 11 3x − 7 + 7 = 11 + 7 3x = 18 x = 6
Check3(6) − 7 = 18 − 7 = 11 ✓

Answer: x = 6

Example 3.5.3

Negative coefficient

ProblemSolve −4x + 3 = −9.
Solve−4x + 3 = −9 −4x + 3 − 3 = −9 − 3 −4x = −12 −4x/(−4) = −12/(−4) x = 3
Check−4(3) + 3 = −12 + 3 = −9 ✓

Answer: x = 3

Example 3.5.4

Division in the original equation

ProblemSolve x/5 − 2 = 6.
ThinkThe variable is divided by 5, and then 2 is subtracted. Undo the subtraction first, then multiply.
Solvex/5 − 2 = 6 x/5 − 2 + 2 = 6 + 2 x/5 = 8 5 · (x/5) = 5 · 8 x = 40
Check40/5 − 2 = 8 − 2 = 6 ✓

Answer: x = 40

Return to the Opening Question

The shipping problem can now be modeled as 8 + 3.50w = 29.00, where w represents the weight in pounds. This is a two-step equation. You will solve it fully in Section 3.13.

Practice 3.5Two-Step Equations

3x + 4 = 19
5x − 7 = 18
−2x + 6 = 14
x/3 + 5 = 11
4x − 1 = −13
x/6 − 3 = 2
10 − x = 4
−x/4 + 3 = 7

Section 3.6

Multi-Step Equations

General Strategy

1. Simplify each side independently (combine like terms).

2. Use inverse operations to isolate the variable term.

3. Solve for the variable.

4. Check by substituting into the original equation.

Example 3.6.1

Combining like terms on one side

ProblemSolve 3x + 2x + 7 = 32.
ThinkCombine 3x and 2x before solving.
Solve3x + 2x + 7 = 32 5x + 7 = 32 5x = 25 x = 5
Check3(5) + 2(5) + 7 = 15 + 10 + 7 = 32 ✓

Answer: x = 5

Example 3.6.2

Multiple constants

ProblemSolve 6x − 4 − 2x + 10 = 22.
ThinkCombine variable terms and constant terms separately on the left side.
Solve6x − 2x − 4 + 10 = 22 4x + 6 = 22 4x = 16 x = 4
Check6(4) − 4 − 2(4) + 10 = 24 − 4 − 8 + 10 = 22 ✓

Answer: x = 4

Practice 3.6Multi-Step Equations

5x + 2x − 3 = 25
4x − 3 + x = 22
8x − 2x − 5 = 13
3x + 2 − x + 6 = 16
7 + 2x − 4 + 5x = 38
−3x + 4x + 9 = 2

Section 3.7

Equations with Parentheses

Parentheses appear in equations when a quantity inside is being multiplied by something outside. The Distributive Property removes the parentheses: a(b + c) = ab + ac. Multiply the factor outside by every term inside, then proceed to solve.

Example 3.7.1

Distributing a positive factor

ProblemSolve 2(x + 3) = 14.
Solve2(x + 3) = 14 2x + 6 = 14 2x = 8 x = 4
Check2(4 + 3) = 2(7) = 14 ✓

Answer: x = 4

Example 3.7.2

Distributing over subtraction

ProblemSolve 3(2x − 5) = 9.
Solve3(2x − 5) = 9 6x − 15 = 9 6x = 24 x = 4
Check3(2(4) − 5) = 3(3) = 9 ✓

Answer: x = 4

Example 3.7.3

Parentheses on both sides

ProblemSolve 4(x + 2) = 2(x + 8).
ThinkDistribute on both sides, then collect like terms.
Solve4(x + 2) = 2(x + 8) 4x + 8 = 2x + 16 4x − 2x = 16 − 8 2x = 8 x = 4
Check4(6) = 24; 2(12) = 24 ✓

Answer: x = 4

Example 3.7.4

Subtracting a parenthetical expression

ProblemSolve 10 − 3(x − 2) = 7.
ThinkDistribute −3 to every term inside. Both terms change sign.
Solve10 − 3(x − 2) = 7 10 − 3x + 6 = 7 16 − 3x = 7 −3x = −9 x = 3
Check10 − 3(3 − 2) = 10 − 3(1) = 7 ✓

Answer: x = 3

Student Warning — Distributing a Negative

When a negative sign precedes a parenthetical expression, every term inside changes sign. A very common mistake is distributing the negative to only the first term. −(x − 4) = −x + 4, not −x − 4. Distribute to every term.

Practice 3.7Equations with Parentheses

3(x + 5) = 24
5(2x − 3) = 25
−2(x + 4) = 10
4(3x − 1) = 44
2(x + 6) = 3(x − 1)
12 − 4(x + 1) = 0
3(2x + 4) = 2(x + 10)
5 − 2(x − 3) = 9

Section 3.8

Equations with Variables on Both Sides

When variable terms appear on both sides of an equation, collect all variable terms on one side and all constant terms on the other. Choose whichever side produces a positive coefficient for the variable to reduce sign errors.

Example 3.8.1

Collecting variable terms on the left

ProblemSolve 7x − 3 = 4x + 9.
ThinkSubtract 4x from both sides to collect variable terms on the left.
Solve7x − 3 = 4x + 9 7x − 4x − 3 = 9 3x − 3 = 9 3x = 12 x = 4
Check7(4) − 3 = 25; 4(4) + 9 = 25 ✓

Answer: x = 4

Example 3.8.2

Variable term larger on the right

ProblemSolve 2x + 11 = 5x − 4.
ThinkThe coefficient of x is larger on the right. Move constants right and variable terms left, or vice versa — both work.
Solve2x + 11 = 5x − 4 11 + 4 = 5x − 2x 15 = 3x x = 5
Check2(5) + 11 = 21; 5(5) − 4 = 21 ✓

Answer: x = 5

Example 3.8.3

Distribution combined with variables on both sides

ProblemSolve 3(x + 4) = x + 20.
Solve3(x + 4) = x + 20 3x + 12 = x + 20 3x − x = 20 − 12 2x = 8 x = 4
Check3(8) = 24; 4 + 20 = 24 ✓

Answer: x = 4

Practice 3.8Variables on Both Sides

6x + 3 = 4x + 11
9x − 5 = 6x + 10
3x + 7 = 7x − 9
5(x + 2) = 2x + 19
4x − 3 = 2(2x − 1) − 1
8x − (3x + 4) = 2x + 14

Section 3.9

Equations with Fractions

Strategy — Clearing Fractions

Multiply both sides of the equation by the least common denominator (LCD) of all fractions in the equation. This eliminates every denominator in one step. Then solve as usual.

Example 3.9.1

Single fraction — solve directly

ProblemSolve x/3 + 4 = 9.
Solvex/3 + 4 = 9 x/3 = 5 x = 15
Check15/3 + 4 = 5 + 4 = 9 ✓

Answer: x = 15

Example 3.9.2

Clearing fractions with the LCD

ProblemSolve x/4 + x/6 = 5.
ThinkDenominators are 4 and 6. LCD = 12. Multiply every term by 12.
Solvex/4 + x/6 = 5 12 · (x/4) + 12 · (x/6) = 12 · 5 3x + 2x = 60 5x = 60 x = 12
Check12/4 + 12/6 = 3 + 2 = 5 ✓

Answer: x = 12

Example 3.9.3

Fraction with a constant in the numerator

ProblemSolve (2x − 1)/3 = 5.
ThinkMultiply both sides by 3 to clear the fraction.
Solve(2x − 1)/3 = 5 3 · (2x − 1)/3 = 3 · 5 2x − 1 = 15 2x = 16 x = 8
Check(2(8) − 1)/3 = 15/3 = 5 ✓

Answer: x = 8

Student Warning

When you multiply both sides by the LCD, you must multiply every term — including terms that are not fractions. A common mistake is to multiply only the fractional terms and leave whole-number terms unchanged.

Practice 3.9Equations with Fractions

x/5 + 3 = 8
2x/3 = 14
x/4 + x/8 = 9
(3x − 2)/5 = 4
(x + 1)/3 = (x − 1)/2
x/6 − x/9 = 2

Section 3.10

Equations with Decimals

Equations with decimals can be solved directly or by clearing the decimals. To clear decimals, multiply both sides by a power of 10 that converts all decimal coefficients to whole numbers. Use 10 for tenths, 100 for hundredths.

Example 3.10.1

Solving directly

ProblemSolve 0.5x = 7.5.
Solve0.5x = 7.5 x = 7.5/0.5 = 15
Check0.5(15) = 7.5 ✓

Answer: x = 15

Example 3.10.2

Clearing decimal tenths (multiply by 10)

ProblemSolve 0.3x + 0.6 = 1.5.
ThinkAll values are in tenths. Multiply every term by 10.
Solve0.3x + 0.6 = 1.5 3x + 6 = 15 3x = 9 x = 3
Check0.3(3) + 0.6 = 0.9 + 0.6 = 1.5 ✓

Answer: x = 3

Example 3.10.3

Clearing decimal hundredths (multiply by 100)

ProblemSolve 0.25x + 1.50 = 4.00.
ThinkValues are in hundredths. Multiply every term by 100.
Solve0.25x + 1.50 = 4.00 25x + 150 = 400 25x = 250 x = 10
Check0.25(10) + 1.50 = 2.50 + 1.50 = 4.00 ✓

Answer: x = 10

Connection — Finance

Decimal equations with hundredths are the standard form of most money problems. Prices, interest rates, and costs are nearly always expressed in tenths or hundredths. The habit of clearing decimals keeps calculations clean and reduces arithmetic errors.

Practice 3.10Equations with Decimals

0.4x = 12
0.5x + 2.5 = 7.5
1.2x − 0.6 = 5.4
0.25x + 0.75 = 2.50
0.1x + 0.3 = 0.7
0.05x + 1.20 = 2.45

Section 3.11

Identity Equations and No-Solution Equations

Most linear equations have exactly one solution. But when variables on both sides eliminate each other entirely, you are left with a statement involving only constants. That constant statement tells you everything.

If the resulting constant statement is true (like 5 = 5 or 0 = 0), the equation is an identity — it is true for every real number. If the resulting statement is false (like 5 = 8 or 0 = −2), the equation is a contradiction — it has no solution.

Example 3.11.1

Identity — all real numbers are solutions

ProblemSolve 3(x + 4) = 3x + 12.
Solve3(x + 4) = 3x + 12 3x + 12 = 3x + 12 0 = 0
Check0 = 0 is always true. Every value of x makes this equation true.

Answer: All real numbers. This is an identity. Solution set = all real numbers.

Example 3.11.2

No solution — contradiction

ProblemSolve 2x + 5 = 2x − 3.
Solve2x + 5 = 2x − 3 5 = −3
Check5 = −3 is never true, for any value of x.

Answer: No solution. Solution set = ∅ (the empty set).

Interpretation

Variable disappears → true statement (0 = 0): all real numbers are solutions (identity).

Variable disappears → false statement (5 = −3): no solution (contradiction).

These are not errors — they are meaningful results about the structure of the equation.

Practice 3.11Identity and No-Solution

Solve each equation. State whether it has one solution, all real numbers, or no solution.

4x + 6 = 4(x + 2) − 2
5x + 3 = 5x − 7
2(3x − 1) = 6x − 2
7x − 4 = 7x + 1
3(x + 5) = 3x + 15
x + 10 = 2x − x + 5

Section 3.12

Writing Equations from Words

Before you can solve a word problem, you must translate the verbal statement into an algebraic equation. This requires careful reading and attention to specific signal words.

Verbal Phrase	Algebraic Meaning
"a number"	`x` (or any variable)
"the sum of x and 5"	`x + 5`
"3 less than x"	`x − 3`
"twice a number"	`2x`
"one-third of a number"	`x/3`
"5 more than three times a number"	`3x + 5`
"a number increased by 4 is 11"	`x + 4 = 11`
"the product of x and 6 is 30"	`6x = 30`
"is", "equals", "is equal to"	`=`

Example 3.12.1

Simple phrase

Problem"Five more than twice a number is 19."
Think"Twice a number" → 2x. "Five more than" → add 5. "Is 19" → equals sign and right side.
Equation2x + 5 = 19

Equation: 2x + 5 = 19

Example 3.12.2

Multi-operation phrase

Problem"Three less than four times a number equals twenty-one."
Think"Four times a number" → 4x. "Three less than" → subtract 3 from 4x, not 3 − 4x.
Equation4x − 3 = 21

Equation: 4x − 3 = 21

Example 3.12.3

Multi-step phrase with division

Problem"A number is divided by 5 and then increased by 3. The result is 11."
Equationx/5 + 3 = 11

Equation: x/5 + 3 = 11

Student Warning — "Less Than"

"Three less than x" means x − 3, not 3 − x. The phrase "less than" describes a reduction from the unknown, not from 3. Always re-read after writing to make sure the subtraction is in the correct order.

Practice 3.12Translation Practice

Write an equation for each sentence. Do not solve.

Eight more than a number is 25.
Four times a number, decreased by 7, is 21.
One-third of a number equals 9.
The sum of three consecutive integers is 60.
Twice the sum of a number and 5 equals 22.
A number divided by 4, then increased by 3, gives 10.
Six less than three times a number equals twice the number plus 5.
The difference between a number and 8 is the same as half of the number.

Section 3.13

Equation Word Problems

Problem-Solving Method

Read — Understand the situation. Identify what is known and unknown.

Represent — Define a variable clearly. Write the equation.

Solve — Use inverse operations to find the variable.

Check — Verify numerically or in context.

Explain — Write a complete sentence with units or context.

Example 3.13.1

The shipping problem (chapter opener)

ReadFlat fee: $8.00. Rate: $3.50 per pound. Total: $29.00. Find: weight.
RepresentLet w = the weight of the package in pounds. 8 + 3.50w = 29
Solve3.50w = 29 − 8 3.50w = 21 w = 21/3.50 = 6
Check8 + 3.50(6) = 8 + 21 = 29 ✓

Explain: The package weighs 6 pounds.

Example 3.13.2

Consecutive integers

ReadTwo consecutive integers differ by 1. Their sum is 47.
RepresentLet the first integer = n. The next = n + 1. n + (n + 1) = 47
Solve2n + 1 = 47 2n = 46 n = 23
Check23 + 24 = 47 ✓

Explain: The two consecutive integers are 23 and 24.

Example 3.13.3

Perimeter problem

ReadPerimeter = 54 cm. Length is 3 cm more than twice the width.
RepresentLet w = width. Then l = 2w + 3. 2(2w + 3) + 2w = 54
Solve4w + 6 + 2w = 54 6w + 6 = 54 6w = 48 w = 8 l = 2(8) + 3 = 19
Check2(19) + 2(8) = 38 + 16 = 54 ✓

Explain: The rectangle is 8 cm wide and 19 cm long.

Example 3.13.4

Rate and time

ReadRate: $12/hr tutoring. Already earned: $42. Total this week: $90. Find: hours in today's session.
RepresentLet h = hours in this session. 42 + 12h = 90
Solve12h = 48 h = 4
Check42 + 12(4) = 42 + 48 = 90 ✓

Explain: The tutoring session was 4 hours long.

Practice 3.13Word Problems

For each problem: define your variable, write an equation, solve, check, and explain in a complete sentence.

A gym membership costs $40 per month plus a one-time enrollment fee of $30. If a member has paid $190 in total, how many months have they been a member?
A student scored 82 on the first test and 79 on the second. What must they score on the third test to have an average of 85?
Two numbers differ by 13. Their sum is 47. Find both numbers.
A plumber charges a flat call-out fee of $55 plus $45 per hour. If a customer paid $190, how many hours did the plumber work?
A runner's total distance for the week is 31 miles. She ran 5 miles each day for the first three days and the same distance on each of the remaining two days. How far did she run each of those last two days?
The perimeter of a rectangle is 72 feet. The length is four feet more than three times the width. Find the length and width.
Two angles are supplementary (they sum to 180°). One angle is 24° more than three times the other. Find both angles.
The sum of four consecutive odd integers is 64. Find the integers. (Hint: Consecutive odd integers differ by 2.)

Section 3.14

Common Equation Mistakes

These mistakes appear frequently in student work. Recognizing them is as important as solving correctly — and recognizing them in your own work before submission is what checking is for.

Mistake 1 — Not operating on both sides

2x + 5 − 5 = 13 → 2x = 13 Correct: 2x + 5 − 5 = 13 − 5, giving 2x = 8, so x = 4. The subtraction must appear on both sides simultaneously.

Mistake 2 — Distributing to only the first term

3(x + 5) = 3x + 5 Correct: 3(x + 5) = 3x + 15. The factor 3 multiplies every term inside the parentheses, not just the first.

Mistake 3 — Sign error distributing a negative

−(x − 4) = −x − 4 Correct: −(x − 4) = −x + 4. The negative distributes to both terms. The minus inside becomes a plus.

Mistake 4 — Forgetting to multiply all terms when clearing fractions

x/3 + 4 = 9 → (×3) → x + 4 = 27

Correct: 3·(x/3) + 3·4 = 3·9  →  x + 12 = 27. Every term must be multiplied by the LCD, including non-fractional terms.

Mistake 5 — Stopping at the variable term without isolating fully

After reaching 3x = 12, writing the answer as "3x = 12" Correct: Continue dividing both sides by 3 to reach x = 4. The goal is always x alone with a coefficient of 1.

Mistake 6 — Not checking the answer

Completing a solution without substituting back to verify

Correct: Always substitute your answer into the original equation. If both sides match, the answer is correct. Checking catches arithmetic errors, sign errors, and distribution mistakes.

Mistake 7 — Misreading "less than"

"Three less than a number" → 3 − x

Correct: "Three less than x" means x − 3. You start with x and remove 3. The English phrase is counterintuitive — the "less" describes the amount removed from the unknown, not a new starting quantity.

Chapter 3 — Back Matter

Chapter Summary

Sec 3.1An equation is a statement that two expressions are equal. A linear equation in one variable contains the variable only to the first power. The equals sign is not a prompt to calculate — it asserts simultaneous equality.
Sec 3.2A solution is a value that makes the equation true. Always verify by substituting into the original equation. Never check with a simplified intermediate form.
Sec 3.3The balance model: both sides of an equation have equal value. The properties of equality allow the same operation on both sides without destroying the balance. Inverse operations — subtraction undoes addition; division undoes multiplication — isolate the variable.
Sec 3.4One-step equations require a single inverse operation. The four types correspond to addition, subtraction, multiplication, and division. Negative coefficients are handled by dividing by the negative number.
Sec 3.5Two-step equations require undoing addition or subtraction first, then multiplication or division. This is the reverse of the order of operations used to build the expression.
Sec 3.6Multi-step equations require simplification — combining like terms — before or alongside inverse operations. Always simplify each side independently first.
Sec 3.7Equations with parentheses require the Distributive Property first. A negative factor outside parentheses changes the sign of every term inside. Distribute before combining or isolating.
Sec 3.8Equations with variables on both sides require collecting all variable terms on one side and all constants on the other. Prefer the side that produces a positive coefficient.
Sec 3.9Equations with fractions are cleared by multiplying every term by the LCD. Every term must be multiplied — not only fractional terms. After clearing, solve as usual.
Sec 3.10Equations with decimals can be cleared by multiplying every term by the appropriate power of 10. Tenths use 10; hundredths use 100. This converts the equation to whole-number coefficients.
Sec 3.11When the variable eliminates and leaves a true constant statement (0 = 0), the equation is an identity with all real numbers as solutions. When it leaves a false statement (5 = −3), there is no solution.
Sec 3.12Verbal descriptions translate to equations using signal words: "sum" → +, "difference" → −, "less than" → reverse order subtraction, "twice" → ×2, "is / equals" → =. Always re-read after translating.
Sec 3.13The Read → Represent → Solve → Check → Explain method organizes word problems. Define the variable clearly, write the equation from the defined variable, solve, verify numerically, and state the answer with context.
Sec 3.14Seven common mistakes: not applying operations to both sides; incomplete distribution; sign errors with negatives; not multiplying all terms when clearing; stopping short of full isolation; skipping the check; misreading "less than."

Key Terms Review

equation

A mathematical statement that two expressions are equal, written with an equals sign.

solution

A value of the variable that makes an equation a true statement.

solution set

The complete collection of all values that make an equation true.

inverse operations

Operations that undo each other. Addition and subtraction are inverse. Multiplication and division are inverse.

properties of equality

Rules that allow the same operation on both sides of an equation without changing its truth.

linear equation

An equation in which the variable appears to the first power only. No x², no radicals, no products of variables.

identity

An equation that is true for every value of the variable. Results in a true constant statement like 0 = 0.

no-solution equation

An equation that is true for no value of the variable. Results in a false constant statement like 5 = −3.

coefficient

The number multiplied by the variable in a term. In −3x, the coefficient is −3. In x alone, it is 1.

LCD

Least Common Denominator — the smallest multiple shared by all denominators in an expression; used to clear fractions.

like terms

Terms with the same variable raised to the same power. Only like terms can be combined by addition or subtraction.

distributive property

The rule a(b + c) = ab + ac. Used to remove parentheses before solving; must be applied to every term inside.

Concept Review Questions

What does it mean for a value to be a solution to an equation?
In the balance model of equations, what does the equals sign represent? What does "performing the same operation on both sides" correspond to?
Why is it necessary to perform the same operation on both sides of an equation?
What are inverse operations? Give two examples of inverse operation pairs.
Explain the general order of operations when solving a two-step equation. Why does that order make sense?
When distributing a factor over parentheses, why must every term inside be multiplied?
What does it mean if, after fully simplifying, you are left with 0 = 0?
What does it mean if you are left with a statement like 5 = 9 after fully simplifying?
Explain how to clear fractions from an equation. What mistake do students often make at this step?
Describe the Read → Represent → Solve → Check → Explain method in your own words. Why does each step matter?

Skill Practice

Section AOne-Step Equations

x + 14 = 23
x − 9 = 5
7x = 63
x/8 = 6
−5x = 40
15 = x − 4

Section BTwo-Step Equations

3x + 4 = 19
5x − 7 = 18
−2x + 6 = 14
x/3 + 5 = 11
4x − 1 = −13
x/6 − 3 = 2

Section CMulti-Step and Parentheses

5x + 2x − 3 = 25
3(x + 4) = 21
2(3x − 1) = 16
5 + 2(x − 3) = 13
4(2x + 3) = 3(2x + 6)
8 − 2(x + 1) = 10

Section DVariables on Both Sides

6x + 3 = 4x + 11
9x − 5 = 6x + 10
3x + 7 = 7x − 9
5(x + 2) = 2x + 19
4x − 3 = 2(2x−1) − 1
8x − (3x+4) = 2x + 14

Section EFractions and Decimals

x/5 + 3 = 8
2x/3 = 14
x/4 + x/8 = 9
(3x − 2)/5 = 4
0.4x = 12
0.5x + 2.5 = 7.5
1.2x − 0.6 = 5.4
0.25x + 0.75 = 2.50

Real-World Applications

Define your variable, write an equation, solve, check, and explain in context.

Finance: A savings account has a balance of $420. Money is withdrawn at a steady rate of $35 per week. After how many weeks will the balance reach $175?
Finance: A store marks up a wholesale item by 40% and sells it for $84. What was the wholesale price? (Hint: Selling price = wholesale price + 40% of wholesale price.)
Science: The formula for converting Celsius to Fahrenheit is F = (9/5)C + 32. If the temperature is 68°F, what is the Celsius temperature?
Science: An experiment starts with 200 mL of water. Water evaporates at 8 mL per hour. After how many hours will only 104 mL remain?
Geometry: A rectangular field has a perimeter of 300 meters. The length is 15 meters more than twice the width. Find the length and width.
Geometry: Two angles are complementary (they sum to 90°). One angle is 12° less than three times the other. Find both angles.
Everyday life: A cell phone plan charges $25 per month plus $0.10 per text message. If a customer's bill was $37.50, how many texts did they send?
Everyday life: A runner's total distance this week is 31 miles. She ran 5 miles each day for the first three days and the same distance on each of the last two days. How far did she run each of those two days?

Error Analysis

Each problem contains a mistake. Identify the error and write the correct solution.

Problem A

3(x + 4) = 21
3x + 4 = 21          ← student's work
3x = 17
x = 17/3

What is the error? What is the correct answer?

Problem B

5x − 3 = 2x + 9
5x − 2x = 9 − 3      ← student's work
3x = 6
x = 2

Check x = 2 in the original. Does it satisfy the equation? If not, find the error and the correct answer.

Problem C

x/4 + 3 = 11
Multiply by 4: x + 3 = 44   ← student's work
x = 41

The student forgot to multiply one term by 4. Identify which term and correct the solution.

Problem D

2x + 8 = 2x − 5
Student's conclusion: "x = 0"

Simplify the equation fully. What does it actually say? What is the correct conclusion?

Problem E

"Nine less than twice a number n equals twelve."
Student writes: 9 − 2n = 12

Is this translation correct? If not, write the correct equation and explain the error.

Chapter 3 Checkpoint

Complete without assistance — assesses all key skills and concepts from Chapter 3

Part 1 — Checking Solutions (no solving required)

Is x = 5 a solution to 3x − 4 = 11? Show your work.
Is x = −2 a solution to 4x + 10 = 2? Show your work.

Part 2 — One-Step and Two-Step Equations

x + 16 = 31
−8x = 56
x/7 = 4
4x − 5 = 23
x/3 + 7 = 12
−3x + 2 = 20

Part 3 — Multi-Step and Parentheses

2(x + 5) = 26
5x − 3x + 8 = 24
3(2x − 4) = 2(x + 6)

Part 4 — Variables on Both Sides

6x − 1 = 4x + 9
3(x + 2) = 2x + 10

Part 5 — Fractions and Decimals

x/6 + 2 = 7
0.4x + 1.2 = 4.0

Part 6 — Identity and No Solution

4x + 6 = 4(x + 2) − 2
5x + 3 = 5x − 7

Part 7 — Word Problem

A gym membership costs $40 per month plus a one-time enrollment fee of $30. A member has paid $190 in total. How many months have they been a member? Use the Read → Represent → Solve → Check → Explain method and show all steps.

Answers may be placed in the answer key.

Bridge Note — Looking Ahead

Chapter 4 and Beyond

The skills built in Chapter 3 are the foundation for almost everything that follows in Algebra I.

Chapter 4 — Inequalities: Inequalities use the same inverse-operation strategy as equations, but produce ranges of solutions rather than single values. One important difference: the inequality symbol reverses when you multiply or divide both sides by a negative number.

Chapter 5 — Functions: Many function problems ask you to evaluate (substitute a value and simplify) or to solve (find which input produces a given output). Both are equation-solving skills.

Chapter 7 — Linear Equations in Two Variables: Every line in the coordinate plane is described by a linear equation. Finding slopes, intercepts, and points on a line all require solving equations.

Chapter 8 — Systems: Solving a system means applying substitution — a direct extension of what you learned here.

Chapters 11–12 — Factoring and Quadratics: The multi-step and distribution skills from this chapter carry directly into solving quadratic equations.

Science, Finance, and Computing: Every physical formula, financial model, and computational algorithm you encounter depends on the same balance-and-inverse-operation reasoning developed in this chapter.

MGU Library Connection

Equation-Solving Strategy Guide — Appendix E of this textbook
Real Number Properties Reference — Appendix B of this textbook
Pre-Algebra Readiness Review — Appendix A of this textbook
Chapter 4: Linear Inequalities in One Variable (next chapter)
Algebra II, Chapter 1: Algebra II Readiness — reviews and extends all equation-solving skills
MGU Dictionary entries: equation, solution, inverse operations, identity, distributive property

Chapter 3:Linear Equationsin One Variable

Purpose and Overview

Learning Objectives

Sections at a Glance

What an Equation Says

The Structure of a Linear Equation

Solutions and Checking

Balance and Inverse Operations

One-Step Equations

Two-Step Equations

Multi-Step Equations

Equations with Parentheses

Equations with Variables on Both Sides

Equations with Fractions

Equations with Decimals

Identity Equations and No-Solution Equations

Writing Equations from Words

Equation Word Problems

Common Equation Mistakes

Chapter Summary

Key Terms Review

Concept Review Questions

Skill Practice

Real-World Applications

Error Analysis

Chapter 3 Checkpoint

Chapter 3:
Linear Equations
in One Variable