SAT — mathMedium7 min read

Solving Multi-Step Linear Equations

Master multi-step linear equations for the Digital SAT. Learn to combine like terms, distribute, and handle variables on both sides.

Tutor AI Team2026-02-25T01:07:10.179Z

Multi-step linear equations are equations that require three or more steps to solve. They appear regularly on the Digital SAT — both as standalone problems and embedded within word problems. The key skills are distributing (expanding brackets), combining like terms, and moving variables from one side to the other.

Unlike simple one- or two-step equations, multi-step equations often have variables and constants on both sides. Mastering these prepares you for systems of equations, function problems, and the more challenging algebra questions that contribute to a high SAT Math score.

Core Concepts

The General Approach

For any multi-step equation, follow this order:

Simplify each side — distribute and combine like terms.
Move variable terms to one side.
Move constant terms to the other side.
Isolate the variable — divide or multiply to get $x$ alone.

Distributing (Expanding Brackets)

When a number is multiplied by a bracketed expression, multiply it by every term inside:

$3(2x - 4) = 6x - 12$

$-(x + 5) = -x - 5$

Be especially careful with a negative sign in front of brackets.

Combining Like Terms

Before moving terms across the equation, simplify each side by combining terms with the same variable:

$4x + 3 - 2x + 7 = (4x - 2x) + (3 + 7) = 2x + 10$

Variables on Both Sides

When $x$ appears on both sides, move one of the variable terms:

Example: Solve $5x + 3 = 2x + 15$

Subtract $2x$ from both sides: $3x + 3 = 15$

Subtract 3: $3x = 12$

Divide by 3: $x = 4$

Tip: Move the smaller variable term to avoid negative coefficients.

Distributing Before Collecting

Example: Solve $2(3x + 1) = 4x + 10$

Distribute: $6x + 2 = 4x + 10$

Subtract $4x$ : $2x + 2 = 10$

Subtract 2: $2x = 8$

Divide by 2: $x = 4$

Clearing Fractions

If the equation contains fractions, multiply every term by the LCD to clear them.

Example: Solve $\frac{x}{3} + \frac{x}{4} = 7$

LCD of 3 and 4 is 12. Multiply every term by 12:

$4x + 3x = 84$

$7x = 84$

$x = 12$

Clearing Decimals

Multiply every term by 10 (or 100) to eliminate decimals:

$0.3x + 1.5 = 2.1$ → multiply by 10 → $3x + 15 = 21$ → $3x = 6$ → $x = 2$

Strategy Tips

Tip 1: Simplify Both Sides First

Always distribute and combine like terms on each side before moving terms across the equals sign.

Tip 2: Choose Which Side to Collect Variables

Collect variables on the side that keeps the coefficient positive. For $3x + 5 = 8x - 10$ , subtract $3x$ from both sides (not $8x$ ) to get $5 = 5x - 10$ , keeping the $x$ term positive.

Tip 3: Clear Fractions Early

Multiplying through by the LCD simplifies the arithmetic and reduces the chance of errors.

Tip 4: Use Backsolving on the SAT

For multiple-choice questions, plug answer choices into the original equation. Start with the middle value to efficiently narrow down the answer.

Tip 5: Be Careful with Distribution

The number-one error source in multi-step equations is incorrect distribution, especially with negatives. $-3(x - 4) = -3x + 12$ , not $-3x - 12$ .

Worked Example: Example 1

Problem

Solve $4(x - 2) + 3 = 2x + 7$

Distribute: $4x - 8 + 3 = 2x + 7$

Combine on left: $4x - 5 = 2x + 7$

Subtract $2x$ : $2x - 5 = 7$

Add 5: $2x = 12$

Divide by 2: $x = 6$

Check: $4(6-2) + 3 = 16 + 3 = 19$ and $2(6) + 7 = 19$ ✓

Solution

Worked Example: Example 2

Problem

Solve $3(2x + 5) = 5(x + 4) + 2$

Distribute both sides: $6x + 15 = 5x + 20 + 2$

Simplify right: $6x + 15 = 5x + 22$

Subtract $5x$ : $x + 15 = 22$

Subtract 15: $x = 7$

Solution

Worked Example: Example 3

Problem

Solve $\frac{2x + 1}{3} = \frac{x - 2}{2} + 1$

Multiply every term by 6 (LCD of 3 and 2):

$2(2x + 1) = 3(x - 2) + 6$

$4x + 2 = 3x - 6 + 6$

$4x + 2 = 3x$

Subtract $3x$ : $x + 2 = 0$

$x = -2$

Solution

Worked Example: SAT Context

Problem

A plumber charges a $40 service fee plus$ 25 per hour. An electrician charges a $60 service fee plus$ 20 per hour. After how many hours will the total charges be equal?

Plumber: $25h + 40$ | Electrician: $20h + 60$

$25h + 40 = 20h + 60$

Subtract $20h$ : $5h + 40 = 60$

Subtract 40: $5h = 20$

$h = 4 \text{ hours}$

Solution

Worked Example: Example 5

Problem

Solve $-2(3 - x) = 4(x + 1) - 2x$

Distribute: $-6 + 2x = 4x + 4 - 2x$

Simplify right: $-6 + 2x = 2x + 4$

Subtract $2x$ : $-6 = 4$

This is a contradiction — no solution exists. The equation has no solution.

Solution

Practice Problems

Problem 1

Solve $5(x + 3) - 2x = 24$ .

Problem 2

Solve $3x - 7 = 2(x + 4)$ .

Problem 3

Solve $\frac{x + 4}{2} = \frac{3x - 2}{4}$ .

Problem 4

Two taxi companies charge different rates. Company A charges $3 plus$ 2 per mile. Company B charges $5 plus$ 1.50 per mile. After how many miles do they charge the same total?

Problem 5

Solve $0.4x + 1.2 = 0.2x + 3.0$ .

Problem 6

Solve $7 - 3(2x - 1) = 4(1 - x)$ .

Want to check your answers and get step-by-step solutions?

Common Mistakes

Incorrect distribution of negatives. $-2(x - 3) = -2x + 6$ , not $-2x - 6$ . Double-check every sign.
Combining unlike terms. You cannot combine $3x$ and $5$ — they are not like terms.
Arithmetic errors when clearing fractions. After multiplying by the LCD, make sure every single term (including constants) gets multiplied.
Forgetting to distribute to all terms inside brackets. In $3(2x + 5)$ , both $2x$ and $5$ are multiplied by 3.
Not simplifying both sides before moving terms. Jumping ahead without simplifying each side first leads to confusion.

Frequently Asked Questions

What if both sides simplify to the same thing?

If you get a true statement like $0 = 0$ or $5 = 5$ , the equation has infinitely many solutions — it's an identity.

What if I get a false statement like $0 = 7$?

The equation has no solution — it's a contradiction. This is tested on the SAT.

Is there a fastest way to solve multi-step equations on the SAT?

For multiple choice, try backsolving (plugging in answer choices). For student-produced response, follow the systematic approach: simplify → collect → isolate.

How do I handle an equation like $2x + 3 = 2x + 3$?

Both sides are identical, so every value of $x$ works. The answer is "infinitely many solutions" or "all real numbers."

Should I always clear fractions first?

Not always, but it usually makes the arithmetic easier. If the fractions are simple, you might solve directly.

Key Takeaways

✓
Follow the four-step process: simplify → move variables → move constants → isolate.
✓
Distribute before combining. Always expand brackets first.
✓
Clear fractions by multiplying through by the LCD for cleaner arithmetic.
✓
Watch for special cases: no solution (contradiction) and infinite solutions (identity).
✓
Check your answer by substituting back into the original equation.
✓
On the SAT, backsolving can be faster than algebraic solving for multiple-choice questions.

Tags:

#sat#math#algebra#equations#multi-step

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Core Concepts

The General Approach

Distributing (Expanding Brackets)

Combining Like Terms

Variables on Both Sides

Distributing Before Collecting

Clearing Fractions

Clearing Decimals

Strategy Tips

Tip 1: Simplify Both Sides First

Tip 2: Choose Which Side to Collect Variables

Tip 3: Clear Fractions Early

Tip 4: Use Backsolving on the SAT

Tip 5: Be Careful with Distribution

Worked Example: Example 1

Worked Example: Example 2

Worked Example: Example 3

Worked Example: SAT Context

Worked Example: Example 5

Practice Problems

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Common Mistakes

Frequently Asked Questions

Key Takeaways

Related Topics

Parallel Structure

Verb Tense and Consistency

Modifier Placement and Dangling Modifiers

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