Solve one- and multi-variable linear equations. - GED Math

SECTION 1

Historical Context & Motivation

Long before modern algebra textbooks existed, people needed to solve problems involving unknown quantities. Ancient merchants calculated how many goods they could trade, builders figured out dimensions for structures, and tax collectors determined fair payments. The idea of using a symbol to stand for an unknown number — and then systematically finding that number — is one of the most powerful ideas in all of mathematics. Understanding linear equations is not just an academic exercise; it is a practical skill used every day in budgeting, cooking, construction, healthcare, and countless other fields.

1800 BCE • Babylonian Clay Tablets

Ancient Babylonians solved linear and quadratic problems on clay tablets using verbal recipes, centuries before symbolic algebra existed.

250 CE • Diophantus of Alexandria

The Greek mathematician Diophantus introduced abbreviations for unknowns and operations, earning the title "Father of Algebra."

820 CE • Al-Khwarizmi's Al-Jabr

The Persian scholar al-Khwarizmi wrote the foundational text on balancing equations. The word "algebra" comes from the Arabic "al-jabr," meaning "restoration" or "completion."

1637 • Descartes' Modern Notation

René Descartes popularized using letters like x, y, and z for unknowns and a, b, and c for constants, giving us the notation we use today.

The central question that all of this history points toward is simple but powerful: if you know a relationship between quantities, how do you figure out the value of the unknown? That is exactly what solving a linear equation does. On the GED Mathematical Reasoning exam, roughly 55% of questions involve algebraic problem solving, and linear equations are the backbone of that content. Mastering this topic puts you in a strong position to succeed.

SECTION 2

Core Principles & Definitions

Before you can solve equations confidently, you need to understand a few foundational ideas. These principles apply whether your equation has one variable or several. Think of them as the rules of the game — once you know them, every equation follows the same playbook.

What Is a Linear Equation?

A linear equation is an equation where every variable is raised to the first power (no exponents, no square roots on variables). Examples: 3x + 5 = 14 or 2a − b = 10.

The Balance Principle

An equation is like a balanced scale. Whatever you do to one side, you must do to the other side to keep it balanced. Add 5 to the left? Add 5 to the right.

Inverse Operations

To undo an operation, use its opposite. Addition undoes subtraction. Multiplication undoes division. These inverse operations are your main tools for isolating a variable.

Isolate the Variable

The goal is always to get the variable alone on one side of the equation. "Solve for x" means rearrange until you have x = (some number or expression).

Combining Like Terms

Before solving, simplify each side by combining like terms — terms that have the same variable raised to the same power. For example, 3x + 2x simplifies to 5x.

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KEY TAKEAWAY

Think of solving an equation like unwrapping a gift. The variable is inside layers of wrapping (addition, multiplication, etc.). You peel off each layer one at a time, using the opposite action, until the variable is revealed. The key rule? Whatever you do to one side, do the exact same thing to the other side.

SECTION 3

Visual Explanation — The Balance Model

The diagram below shows how solving the equation 2x + 3 = 9 works as a series of balancing steps. Each transformation keeps both sides equal while peeling away layers until x stands alone.

Each row represents the equation at a different stage. The triangle is the fulcrum of the balance. Notice how subtracting 3 from both sides (Step 1) and dividing both sides by 2 (Step 2) keeps the scale balanced at every stage, leading to the solution x = 3.

This visual reinforces the most important rule in equation solving: every operation you perform must be done to both sides. If you subtract 3 from the left but forget to subtract 3 from the right, the scale tips and the equation is no longer true. Keep this image in mind whenever you work through a problem.

SECTION 4

Mathematical Framework

Let's formalize the techniques you will use on the GED. There are two main categories: one-variable linear equations (like 3x − 7 = 14) and multi-variable equations (like solving d = rt for t). Both use the same inverse-operation strategy.

One-Variable Equations

STANDARD FORM

ax + b = c

where a is the coefficient of x, b is the constant on the variable's side, and c is the constant on the other side. Solve by: (1) subtract b from both sides → ax = c − b, then (2) divide both sides by a → x = (c − b) ÷ a.

Equations with Variables on Both Sides

VARIABLES ON BOTH SIDES

ax + b = cx + d

First, move all variable terms to one side: subtract cx from both sides to get (a − c)x + b = d. Then subtract b from both sides: (a − c)x = d − b. Finally divide by (a − c).

Multi-Variable (Literal) Equations

REARRANGING A FORMULA

d = rt → t = d ÷ r

When a formula has multiple variables, you "solve for" one of them by treating every other variable like a number. Here, to solve for t, divide both sides by r. The GED formula sheet gives you formulas like this and may ask you to rearrange them.

PERIMETER FORMULA REARRANGED

P = 2l + 2w → w = (P − 2l) ÷ 2

To solve the perimeter formula for w: subtract 2l from both sides to get P − 2l = 2w, then divide both sides by 2. This is exactly the kind of multi-variable rearrangement the GED tests.

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GED Tip: Distributive Property

Many GED equations include parentheses, like 3(x + 4) = 21. Before you isolate the variable, distribute: 3 × x + 3 × 4 = 3x + 12. Then solve 3x + 12 = 21 using the standard steps. Alternatively, you can divide both sides by 3 first to get x + 4 = 7, then subtract 4. Either path gives x = 3.

SECTION 5

Step-by-Step Solving Strategy

Whether you are solving a simple one-step equation or a complex multi-variable formula, the same general strategy applies. The flowchart below lays out the decision-making process. Following these steps in order will keep you organized and prevent mistakes.

This five-step flowchart works for one-variable equations and for rearranging multi-variable formulas. Steps 2 and 3 may swap order depending on the equation, and simpler equations may skip steps that don't apply. Always end with Step 5: Check — substituting your answer back in catches arithmetic errors before you submit.

Equation Type	Example	Key Steps
One-step	x + 7 = 12	One inverse operation: subtract 7
Two-step	3x − 5 = 10	Add 5 to both sides, then divide by 3
Variables both sides	5x + 2 = 3x + 10	Subtract 3x from both sides, then solve the two-step equation
With parentheses	2(x − 4) = 14	Distribute first (or divide by 2), then solve
Multi-variable (literal)	A = ½bh, solve for h	Treat other variables as constants; multiply by 2, divide by b

Common equation types you will see on the GED

SECTION 6

Worked Examples

Example 1: One-Variable Equation with Variables on Both Sides

A phone plan charges $25 per month plus $0.10 per text message. Another plan charges $15 per month plus $0.20 per text. After how many texts do both plans cost the same?

Solve: 25 + 0.10t = 15 + 0.20t

1
Step 1 — Set Up the Equation

Let t represent the number of text messages. Plan A costs 25 + 0.10t dollars. Plan B costs 15 + 0.20t dollars. Set them equal.

25 + 0.10t = 15 + 0.20t
2
Step 2 — Move Variable Terms to One Side

Subtract 0.10t from both sides so the variable terms are on the right.

25 = 15 + 0.10t
3
Step 3 — Move Constants to the Other Side

Subtract 15 from both sides to isolate the term with t.

10 = 0.10t
4
Step 4 — Divide by the Coefficient

Divide both sides by 0.10.

t = 100
5
Step 5 — Check

Plan A: 25 + 0.10(100) = 25 + 10 = $35. Plan B: 15 + 0.20(100) = 15 + 20 = $35. Both equal $35. ✓

t = 100 text messages

Example 2: Multi-Variable Equation (Rearranging a Formula)

The formula for the area of a trapezoid is on the GED formula sheet: A = ½h(b₁ + b₂). Suppose you know the area, the height, and one base, and need to find the other base. Solve for b₂.

Solve A = ½h(b₁ + b₂) for b₂

1
Step 1 — Eliminate the Fraction

Multiply both sides by 2 to remove the ½.

2A = h(b₁ + b₂)
2
Step 2 — Divide by h

Divide both sides by h to unwrap the parentheses.

2A ÷ h = b₁ + b₂
3
Step 3 — Isolate b₂

Subtract b₁ from both sides.

b₂ = (2A ÷ h) − b₁
4
Step 4 — Verify with Numbers

If A = 30, h = 5, and b₁ = 4: b₂ = (2 × 30 ÷ 5) − 4 = 12 − 4 = 8. Check: ½ × 5 × (4 + 8) = ½ × 5 × 12 = 30. ✓

b₂ = 8 ✓

SECTION 7

Common Errors & How to Avoid Them

Knowing the right steps is only half the battle. On test day, it's equally important to recognize the mistakes that trip people up. The table below highlights the most frequent errors and how to avoid them.

Common Error	What Goes Wrong	How to Fix It
Forgetting to distribute	Writing 3(x + 2) as 3x + 2 instead of 3x + 6	Multiply the number outside by every term inside the parentheses
Sign errors	Moving −5 to the other side but writing −5 instead of +5	Remember: moving a term flips its sign. Subtract becomes add, and vice versa
Dividing only one side	Dividing just the variable side by the coefficient	Always apply the operation to the entire other side. Circle both sides after each step
Combining unlike terms	Adding 3x + 4 to get 7x	Only combine terms with the same variable. 3x and 4 are not like terms
Skipping the check	Submitting without verifying	Plug your answer back into the original equation. If both sides are equal, you're good

Top 5 mistakes on linear equation problems

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EXAM STRATEGY

On the no-calculator portion of the GED (Part 1), you'll likely see one or two linear equations with small, friendly numbers. Work carefully and always check your answer by substituting back in. On Part 2, use the TI-30XS calculator for decimal or fraction arithmetic, but set up the algebra by hand first.

SECTION 8

Connection to Advanced Topics

Linear equations are the foundation for more complex algebra that also appears on the GED. Once you are comfortable solving single linear equations, you are ready to tackle systems of two equations, linear inequalities, and linear functions (graphing). The table below compares what you've learned here to these next-level topics.

This Lesson	Next Step on the GED
Solve one equation with one variable: 3x + 5 = 20	Solve a system of two equations with two variables: y = 2x + 1 and y = −x + 7
Use the equals sign (=) to find an exact answer	Use inequality symbols (<, >, ≤, ≥) to find a range of solutions
Rearrange d = rt to solve for t	Graph the relationship y = mx + b on a coordinate plane
Variables to the first power only	Quadratic equations where variables appear squared: x² + 5x + 6 = 0

How this lesson connects to other GED topics

The good news is that every single one of these advanced topics builds directly on the balance principle and inverse operations you've practiced here. If you can solve a linear equation, you already have the core skills for everything else in the algebra section of the GED.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student is solving the equation 4x − 9 = 15. As the first step, the student adds 9 to both sides. Which of the following best explains why this step is valid?

A. Adding 9 makes the left side equal to zero.
B. Adding the same value to both sides of an equation keeps the equation balanced.
C. Adding 9 is the same as multiplying both sides by 9.
D. You must always add a number before dividing.

Show answer (B)

The correct answer is B. The balance principle states that performing the same operation on both sides of an equation preserves equality. Adding 9 to both sides undoes the subtraction of 9 on the left. Choice A is wrong because adding 9 to the left gives 4x, not zero. Choice C is wrong because addition and multiplication are different operations. Choice D is wrong because there is no fixed rule about the order; you choose addition here because it is the inverse of the subtraction in the equation.

Question 2: BASIC CALCULATION

Solve for x: 5x + 12 = 47

A. x = 7
B. x = 9.4
C. x = 11.8
D. x = 35

Show answer (A)

The correct answer is A. Subtract 12 from both sides: 5x = 35. Then divide both sides by 5: x = 7. Check: 5(7) + 12 = 35 + 12 = 47. ✓ Choice B (9.4) comes from dividing 47 by 5 without subtracting 12 first. Choice C (11.8) comes from dividing 47 + 12 = 59 by 5, which is adding instead of subtracting. Choice D (35) is the result after step one (5x = 35) but forgets to divide by 5.

Question 3: INTERMEDIATE

A gym charges a one-time registration fee of $40 plus $25 per month. Another gym has no registration fee but charges $35 per month. Marcus sets up the equation 40 + 25m = 35m to find after how many months the total costs are equal. What is the value of m?

A. m = 2
B. m = 4
C. m = 6
D. m = 10

Show answer (B)

The correct answer is B. Subtract 25m from both sides: 40 = 10m. Divide both sides by 10: m = 4. Check: Gym 1 costs 40 + 25(4) = $140. Gym 2 costs 35(4) = $140. ✓ Choice A (m = 2) likely comes from dividing 40 by 20 (an incorrect coefficient difference). Choice C (m = 6) would mean Gym 1 costs $190 and Gym 2 costs $210 — not equal. Choice D (m = 10) may result from dividing 40 + 60 incorrectly.

Question 4: APPLIED

The formula for converting temperature from Celsius to Fahrenheit is F = (9/5)C + 32. A recipe says to bake at 350°F. Solve this formula for C and then find the Celsius equivalent of 350°F. What is C, rounded to the nearest whole number?

A. C ≈ 162
B. C ≈ 177
C. C ≈ 194
D. C ≈ 212

Show answer (B)

The correct answer is B. First, rearrange the formula: subtract 32 from both sides → F − 32 = (9/5)C. Then multiply both sides by 5/9 → C = 5(F − 32)/9. Plug in F = 350: C = 5(350 − 32)/9 = 5(318)/9 = 1590/9 ≈ 176.67, which rounds to 177. Choice A (162) results from subtracting 32 but then dividing by 9/5 incorrectly. Choice C (194) may come from a computation error such as using 350 instead of 318. Choice D (212) confuses 350°F with the boiling point conversion.

Question 5: CRITICAL THINKING

3 pts

The perimeter of a rectangle is 58 inches. The length is 5 inches more than twice the width. Using the perimeter formula P = 2l + 2w, find both the width and the length of the rectangle. Show your work and explain each step.

Show answer

Rubric (3 points): 1 point — Correctly sets up the system of relationships: l = 2w + 5 and 2l + 2w = 58. 1 point — Substitutes l = 2w + 5 into the perimeter equation and solves correctly: 2(2w + 5) + 2w = 58 → 4w + 10 + 2w = 58 → 6w + 10 = 58 → 6w = 48 → w = 8 inches. 1 point — Finds the length using l = 2(8) + 5 = 21 inches and verifies: 2(21) + 2(8) = 42 + 16 = 58. ✓ Full solution: Let w = width. The length is l = 2w + 5. Substitute into P = 2l + 2w: 58 = 2(2w + 5) + 2w. Distribute: 58 = 4w + 10 + 2w. Combine like terms: 58 = 6w + 10. Subtract 10: 48 = 6w. Divide by 6: w = 8 inches. Then l = 2(8) + 5 = 21 inches. Check: 2(21) + 2(8) = 42 + 16 = 58 inches. ✓

SUMMARY

Lesson Summary

Summary

A linear equation contains variables raised only to the first power. To solve one, use inverse operations (addition/subtraction, multiplication/division) to isolate the variable, always applying the balance principle — do the same thing to both sides. The five-step strategy is: (1) simplify each side, (2) move variable terms to one side, (3) move constants to the other, (4) divide by the coefficient, and (5) check by substituting back into the original equation.

For multi-variable (literal) equations, the process is identical — treat every variable except the one you're solving for as if it were a constant, and use the same inverse operations. Watch out for common errors like forgetting to distribute, sign errors, and combining unlike terms. These skills form the foundation for systems of equations, inequalities, and graphing linear functions — all key topics on the GED.