Skill: multi-step-equations

Multi-Step Equations

When x appears on both sides — or hides inside parentheses — collect, combine, and conquer. And sometimes the answer is that there is no answer.

1 Understand

The core idea in plain language.

A multi-step equation has x on both sides, like 2x + 3 = x + 7. The plan: collect the x-terms on one side (subtract x from both sides → x + 3 = 7), then finish as a two-step equation → x = 4.

If parentheses appear, distribute first: 2(x + 1) = 3x − 1 becomes 2x + 2 = 3x − 1. Only after the parentheses are gone do you collect x-terms.

Two special endings exist. 3x + 5 = 3x + 9 collapses to 0 = 4 — never true, so there is no solution. 2x + 4 = 2(x + 2) collapses to 0 = 0 — always true, so there are infinitely many solutions.

Fractions and decimals need no new rules — just clear them when convenient. In x/2 + 3 = 5, multiply everything by 2 first: x + 6 = 10. In 0.5x + 4 = 2.5x, subtract 0.5x: 4 = 2x.

2 See it

Diagrams that make the idea visual.

Collect x on one side

Subtract the smaller x-term from both sides so the coefficient stays positive.

2x + 3 = x + 7 → x + 3 = 7 → x = 4.

Collect the variable terms first; the rest is a two-step equation.

The two special endings

Sometimes collecting x makes every x vanish. What is left decides everything.

0 = 4 is impossible → no solution. 0 = 0 is always true → infinitely many solutions.

False statement: no solution. True statement: every x works.

3 The key rule

The multi-step order

distribute → collect x-terms → two-step finish

1) Distribute to clear parentheses. 2) Add/subtract to gather all x-terms on one side and all numbers on the other. 3) Combine like terms. 4) Divide by the x-coefficient. 5) If x vanishes: false statement = no solution, true statement = infinitely many.

4 Worked examples

Follow each step. The pattern is always the same.

Example 1 Variables on both sides

Subtract x from both sides: x + 3 = 7.
Subtract 3: x = 4.

2x + 3 = x + 7 → x = 4

Check: 2(4) + 3 = 11 and 4 + 7 = 11. Correct.

Example 2 Distribute first

Distribute: 2(x − 3) = 2x − 6, so 2x − 6 = x + 1.
Subtract x: x − 6 = 1. Add 6: x = 7.

2(x − 3) = x + 1 → x = 7

Check: 2(7 − 3) = 8 and 7 + 1 = 8. Correct.

Example 3 No solution

Subtract 3x from both sides: 5 = 9.
False — the x-terms were identical, so no x can fix it.

3x + 5 = 3x + 9 → no solution

Check: Both sides grow at the same rate; the +5 vs +9 gap never closes.

Example 4 Infinitely many solutions

Distribute the right side: 2(x + 2) = 2x + 4.
Both sides are identical: 0 = 0. Every x works.

2x + 4 = 2(x + 2) → infinitely many solutions

Check: Try x = 10: 24 = 24. Try x = −3: −2 = −2. Always equal.

5 Common mistakes

These errors show up on almost every quiz. Spot them now and they will never cost you points.

1. Forgetting to distribute to every term

Wrong

2(x − 3) = x + 1 → “2x − 3 = x + 1” — the −3 never got doubled.

Right

2(x − 3) = 2x − 6. Every term inside meets the 2.

Distribute across ALL terms in the parentheses.

2. Collecting x on the side that makes negatives

Wrong

2x + 3 = x + 7 → −x: “x + 3 = 7” is fine, but subtracting 2x gives “3 = −x + 7” — messier.

Right

Subtract the smaller x-term so the coefficient stays positive.

Work with positive coefficients whenever you have the choice.

3. Calling 0 = 0 “no solution”

Wrong

“0 = 0, so there is no solution.”

Right

0 = 0 is true — every x satisfies it. That is infinitely many solutions.

False statement → no solution. True statement → infinitely many.

6 Key vocabulary

Say these words like you mean them.

Collect like terms: Gather variable terms on one side, constants on the other.
Equivalent equation: A simpler equation with the same solution.
No solution: No value of x makes the equation true (ends in a false statement like 0 = 4).
Infinitely many solutions: Every value of x works (ends in a true statement like 0 = 0).
Identity: An equation true for all x, e.g. 2(x + 2) = 2x + 4.

7 Quick check

Try each one on paper first, then reveal the answer.

1. Solve: 4x − 1 = 2x + 7.

Answer

Subtract 2x: 2x − 1 = 7; add 1: 2x = 8; x = 4.

2. Solve: 5x + 2 = 3x + 10.

Answer

Subtract 3x: 2x + 2 = 10; subtract 2: 2x = 8; x = 4.

3. Solve: 3(x + 2) = 2x + 9.

Answer

Distribute: 3x + 6 = 2x + 9; subtract 2x: x + 6 = 9; x = 3.

4. How many solutions does 4x + 3 = 4x + 8 have?

Answer

Subtract 4x: 3 = 8 — false. No solution.

Key points to remember

Distribute first, then collect x-terms on one side.
Subtract the smaller x-term to keep coefficients positive.
0 = false statement → no solution; 0 = true statement → infinitely many.
Fractions: multiply by the denominator to clear them early.
Always substitute to verify — both sides must agree.

Practice problems Try the tool Explore the lab