Multi-Step Linear Equations
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Multi-Step Linear Equation
A multi-step linear equation is an equation that requires two or more operations to isolate the variable. These equations may involve combining like terms, using the distributive property, and performing inverse operations on both sides.
Solving multi-step linear equations is a foundational algebra skill that builds on single-step equations. The goal is always the same: isolate the variable on one side of the equation.
General Strategy for Solving Multi-Step Equations
- Simplify each side - Distribute if needed and combine like terms
- Collect variable terms - Get all variable terms on one side
- Collect constant terms - Get all constants on the other side
- Isolate the variable - Divide or multiply to solve for the variable
- Check your solution - Substitute back into the original equation
Key Principle: Balance
Whatever operation you perform on one side of an equation, you must perform the same operation on the other side. This maintains the equality and is the foundation of solving all equations.
Types of Multi-Step Equations
| Type | Example | Key Steps |
|---|---|---|
| Combining Like Terms | 3x + 2x - 5 = 20 | Combine 3x + 2x first |
| Distributive Property | 2(x + 4) = 14 | Distribute the 2 first |
| Variables on Both Sides | 5x - 3 = 2x + 9 | Move variables to one side |
| Fractions | (x + 2)/3 = 5 | Multiply to clear fractions |
SAT/ACT Connection
Multi-step linear equations appear frequently on standardized tests. The SAT Algebra domain and ACT Math section both test your ability to solve these equations efficiently. Practice recognizing patterns to save time on test day.
Examples
Work through these examples to see the step-by-step process for solving multi-step equations.
Example 1: Combining Like Terms
Problem: Solve 4x + 7 - x = 19
Step 1: Combine like terms on the left side
4x - x + 7 = 19
3x + 7 = 19
Step 2: Subtract 7 from both sides
3x = 12
Step 3: Divide both sides by 3
x = 4
Check: 4(4) + 7 - 4 = 16 + 7 - 4 = 19 ✓
Example 2: Using the Distributive Property
Problem: Solve 3(2x - 5) = 21
Step 1: Distribute 3 to both terms in parentheses
6x - 15 = 21
Step 2: Add 15 to both sides
6x = 36
Step 3: Divide both sides by 6
x = 6
Check: 3(2(6) - 5) = 3(12 - 5) = 3(7) = 21 ✓
Example 3: Variables on Both Sides
Problem: Solve 7x - 4 = 3x + 12
Step 1: Subtract 3x from both sides to collect variables
4x - 4 = 12
Step 2: Add 4 to both sides
4x = 16
Step 3: Divide both sides by 4
x = 4
Check: 7(4) - 4 = 24 and 3(4) + 12 = 24 ✓
Example 4: Distributive Property on Both Sides
Problem: Solve 2(x + 3) = 4(x - 1)
Step 1: Distribute on both sides
2x + 6 = 4x - 4
Step 2: Subtract 2x from both sides
6 = 2x - 4
Step 3: Add 4 to both sides
10 = 2x
Step 4: Divide both sides by 2
x = 5
Check: 2(5 + 3) = 2(8) = 16 and 4(5 - 1) = 4(4) = 16 ✓
Example 5: Clearing Fractions
Problem: Solve (2x + 1)/4 = 3
Step 1: Multiply both sides by 4 to clear the fraction
2x + 1 = 12
Step 2: Subtract 1 from both sides
2x = 11
Step 3: Divide both sides by 2
x = 5.5 or x = 11/2
Check: (2(5.5) + 1)/4 = (11 + 1)/4 = 12/4 = 3 ✓
Practice
Try these problems on your own. Choose the best answer for each question.
1. Solve: 5x + 3 = 28
A) x = 4 B) x = 5 C) x = 6 D) x = 7
2. Solve: 2(x - 4) = 18
A) x = 7 B) x = 9 C) x = 11 D) x = 13
3. Solve: 6x - 5 = 4x + 9
A) x = 2 B) x = 5 C) x = 7 D) x = 14
4. Solve: 3(x + 2) - x = 14
A) x = 2 B) x = 4 C) x = 6 D) x = 8
5. Solve: 4x + 7 = 2x + 19
A) x = 3 B) x = 6 C) x = 9 D) x = 12
6. Solve: 5(2x - 1) = 35
A) x = 3 B) x = 4 C) x = 5 D) x = 6
7. Solve: 8x - 3(x + 4) = 13
A) x = 3 B) x = 5 C) x = 7 D) x = 9
8. Solve: (3x - 6)/3 = 7
A) x = 7 B) x = 9 C) x = 11 D) x = 13
9. Solve: 2(3x + 4) = 5(x + 2)
A) x = 1 B) x = 2 C) x = 3 D) x = 4
10. Solve: 9 - 2(x - 1) = 3x + 1
A) x = 1 B) x = 2 C) x = 3 D) x = 4
Click to reveal answers
- B) x = 5 - 5x + 3 = 28; 5x = 25; x = 5
- D) x = 13 - 2x - 8 = 18; 2x = 26; x = 13
- C) x = 7 - 2x = 14; x = 7
- B) x = 4 - 3x + 6 - x = 14; 2x + 6 = 14; 2x = 8; x = 4
- B) x = 6 - 2x = 12; x = 6
- B) x = 4 - 10x - 5 = 35; 10x = 40; x = 4
- B) x = 5 - 8x - 3x - 12 = 13; 5x = 25; x = 5
- B) x = 9 - 3x - 6 = 21; 3x = 27; x = 9
- B) x = 2 - 6x + 8 = 5x + 10; x = 2
- B) x = 2 - 9 - 2x + 2 = 3x + 1; 11 - 2x = 3x + 1; 10 = 5x; x = 2
Check Your Understanding
Answer these reflection questions to deepen your understanding.
1. Why is it important to perform the same operation on both sides of an equation?
Reveal Answer
An equation represents a balance where both sides are equal. If you perform an operation on only one side, you break this balance and the equation is no longer true. By performing the same operation on both sides, you maintain the equality while transforming the equation into a simpler form.
2. When solving equations with variables on both sides, why do we typically move variable terms to one side first?
Reveal Answer
Moving variable terms to one side first allows us to combine them into a single term, making it easier to isolate the variable. This simplifies the equation and reduces the number of steps needed to find the solution. You can actually do the steps in different orders, but this approach is usually most efficient.
3. What is the advantage of checking your solution by substituting it back into the original equation?
Reveal Answer
Checking your solution verifies that you solved the equation correctly. Arithmetic errors can occur during the solving process, and substitution catches these mistakes. It also confirms that your answer makes mathematical sense. On standardized tests, checking can help you avoid losing points on problems you could have solved correctly.
4. Explain the distributive property and why it's essential for solving multi-step equations.
Reveal Answer
The distributive property states that a(b + c) = ab + ac. It allows us to remove parentheses by multiplying the outside term by each term inside the parentheses. This is essential because we cannot combine like terms or isolate variables until all parentheses are eliminated and terms are in their simplest form.
🚀 Next Steps
- Review any concepts that felt challenging
- Move on to the next lesson when ready
- Return to practice problems periodically for review