Solve Two-Step
Linear Equations
Master equations of the form \( px + q = r \) by learning to undo operations in the right order and keep both sides balanced.
7th Grade Math
What Is a Two-Step Equation?
A two-step equation is an equation that requires exactly two inverse operations to isolate the variable. The general forms are:
Think of it like a locked box: you need two keys (two operations) to open it and find \( x \).
The Balance Model
An equation is like a balance scale. Whatever you do to one side, you must do to the other to keep it balanced.
One Operation at a Time
Every two-step equation involves two operations applied to the variable. To solve, undo them in reverse order: peel off the outer operation first, then the inner one.
Step 1
Undo addition or subtraction (the operation farthest from \( x \)).
Step 2
Undo multiplication or division (the operation closest to \( x \)).
Two Forms of Two-Step Equations
| \( px + q = r \) | \( p(x + q) = r \) | |
|---|---|---|
| Example | \( 3x + 5 = 20 \) | \( 3(x + 5) = 21 \) |
| Step 1 | Subtract 5 | Divide by 3 |
| Step 2 | Divide by 3 | Subtract 5 |
| Key Idea | Undo add, then multiply | Undo multiply, then add |
One-Step vs. Two-Step
One-Step
\( x + 5 = 12 \)
Only one operation to undo.
Subtract 5: \( x = 7 \)
Only one operation to undo.
Subtract 5: \( x = 7 \)
Two-Step
\( 3x + 5 = 20 \)
Two operations to undo.
Subtract 5, then divide by 3: \( x = 5 \)
Two operations to undo.
Subtract 5, then divide by 3: \( x = 5 \)
Inverse Operations
Each operation has a partner that undoes it. These are called inverse operations:
Addition ↔ Subtraction
\( +5 \) is undone by \( -5 \)
Multiplication ↔ Division
\( \times 3 \) is undone by \( \div 3 \)
How to Solve Step by Step
1
Identify the variable term and the two operations applied to it.
2
Undo addition or subtraction first: add or subtract the same number on both sides.
3
Undo multiplication or division next: multiply or divide both sides by the same number.
4
Check your answer: substitute back into the original equation to verify.
Example 1: Solve \( 2x + 7 = 15 \)
Start:
\( 2x + 7 = 15 \)
Step 1:
\( 2x + 7 \color{#fbbf24}{- 7} = 15 \color{#fbbf24}{- 7} \) → \( 2x = 8 \)
Step 2:
\( \dfrac{2x}{\color{#4ade80}{2}} = \dfrac{8}{\color{#4ade80}{2}} \) → \( x = 4 \)
Example 2: Solve \( 5x - 3 = 22 \)
Start:
\( 5x - 3 = 22 \)
Step 1:
\( 5x - 3 \color{#fbbf24}{+ 3} = 22 \color{#fbbf24}{+ 3} \) → \( 5x = 25 \)
Step 2:
\( \dfrac{5x}{\color{#4ade80}{5}} = \dfrac{25}{\color{#4ade80}{5}} \) → \( x = 5 \)
Check:
\( 5(5) - 3 = 25 - 3 = 22 \) ✓
Example 3: Solve \( 4(x + 2) = 36 \)
Start:
\( 4(x + 2) = 36 \)
Step 1:
\( \dfrac{4(x+2)}{\color{#818cf8}{4}} = \dfrac{36}{\color{#818cf8}{4}} \) → \( x + 2 = 9 \)
Step 2:
\( x + 2 \color{#f87171}{- 2} = 9 \color{#f87171}{- 2} \) → \( x = 7 \)
Check:
\( 4(7 + 2) = 4(9) = 36 \) ✓
Example 4: Solve \( \frac{x}{3} + 4 = 10 \)
Start:
\( \dfrac{x}{3} + 4 = 10 \)
Step 1:
\( \dfrac{x}{3} + 4 \color{#fbbf24}{- 4} = 10 \color{#fbbf24}{- 4} \) → \( \dfrac{x}{3} = 6 \)
Step 2:
\( \dfrac{x}{3} \color{#4ade80}{\times 3} = 6 \color{#4ade80}{\times 3} \) → \( x = 18 \)
Check:
\( \dfrac{18}{3} + 4 = 6 + 4 = 10 \) ✓
From Words to Equations
Many real-world problems translate into two-step equations. Follow this process:
1
Define the variable: What does \( x \) represent?
2
Write the equation: Translate the sentence into math symbols.
3
Solve: Use the two-step method.
4
Interpret and check: Does the answer make sense in context?
Word Problem Example
Define:
Let \( x \) = number of months
Equation:
\( 25x + 40 = 190 \)
Step 1:
\( 25x + 40 \color{#fbbf24}{- 40} = 190 \color{#fbbf24}{- 40} \) → \( 25x = 150 \)
Step 2:
\( \dfrac{25x}{\color{#4ade80}{25}} = \dfrac{150}{\color{#4ade80}{25}} \) → \( x = 6 \)
Answer:
You paid for 6 months. Check: \( 25(6) + 40 = 150 + 40 = 190 \) ✓
Common Mistakes
✕ Wrong Order
Dividing before subtracting in \( px + q = r \). Always undo addition/subtraction first.
✕ One Side Only
Subtracting from one side but forgetting the other. Both sides must stay equal.
✕ Sign Errors
Writing \( -3 + 3 = 6 \) instead of \( 0 \). Track positive and negative signs carefully.
Quick Reference Summary
| Step | \( px + q = r \) | \( p(x + q) = r \) |
|---|---|---|
| Step 1 | Subtract (or add) \( q \) | Divide by \( p \) |
| Step 2 | Divide (or multiply) by \( p \) | Subtract (or add) \( q \) |
| Result | \( x = \dfrac{r - q}{p} \) | \( x = \dfrac{r}{p} - q \) |
| Always | Check by substituting back into the original equation | |
Question 1 of 5
Solve: \( 4x + 9 = 29 \)
Subtract 9 from both sides: \( 4x = 20 \). Then divide by 4: \( x = 5 \). Check: \( 4(5) + 9 = 20 + 9 = 29 \) ✓
Question 2 of 5
Solve: \( 3(x + 4) = 27 \)
Divide both sides by 3: \( x + 4 = 9 \). Then subtract 4: \( x = 5 \). Check: \( 3(5 + 4) = 3(9) = 27 \) ✓
Question 3 of 5
Solve: \( \dfrac{x}{2} - 6 = 5 \)
Add 6 to both sides: \( \frac{x}{2} = 11 \). Multiply both sides by 2: \( x = 22 \). Check: \( \frac{22}{2} - 6 = 11 - 6 = 5 \) ✓
Question 4 of 5
A pizza shop charges \( \$8 \) per pizza plus a \( \$5 \) delivery fee. If you spent \( \$37 \), how many pizzas did you order?
Equation: \( 8x + 5 = 37 \). Subtract 5: \( 8x = 32 \). Divide by 8: \( x = 4 \). You ordered 4 pizzas. Check: \( 8(4) + 5 = 32 + 5 = 37 \) ✓
Question 5 of 5
What is the first step to solve \( 6x - 11 = 25 \)?
The equation has \( -11 \), so undo it by adding 11 to both sides first. This gives \( 6x = 36 \), and then you divide by 6 to get \( x = 6 \).