# How to solve multi-step equations

Canonical: https://duckyhelper.com/learn/algebra-1/solving-equations/
Updated: 2026-10-01

To solve a multi-step equation, undo what was done to the variable. Clear parentheses and fractions first, then combine like terms on each side. Next, move the variable terms to one side and the numbers to the other, and divide by the coefficient. Always check by putting your answer back into the original equation. If the variable cancels out, there is either no solution or every number works.

## The key idea

An equation says two things are equal. You can do anything you like to it, as long as you do the same thing to both sides. Each move uses an inverse operation: subtracting undoes adding, and dividing undoes multiplying.

$$
2x + 5 = 17 \;\Longrightarrow\; 2x = 12 \;\Longrightarrow\; x = 6
$$

A longer equation needs more moves, but they always come in the same order:

1. Distribute to clear parentheses.
2. Clear fractions or decimals by multiplying **every** term by the same number.
3. Combine like terms on each side.
4. Move all variable terms to one side.
5. Move all plain numbers to the other side.
6. Divide by the number in front of the variable.
7. Check your answer in the original equation.

## Worked examples

**Example 1: variables on both sides**

Problem: Solve \(5x - 7 = 2x + 11\).

1. Subtract \(2x\) from both sides so the variable is only on the left.

   $$
   3x - 7 = 11
   $$
2. Add 7 to both sides.

   $$
   3x = 18
   $$
3. Divide both sides by 3.

   $$
   x = 6
   $$
4. Check: both sides give 23.

   $$
   5(6) - 7 = 23 = 2(6) + 11
   $$

Answer: \(x = 6\)

**Example 2: distributing a negative**

Problem: Solve \(4 - 3(2x - 5) = 2(x + 1) - 7\).

1. Distribute on both sides. The \(-3\) multiplies both terms, so \(-3 \cdot -5 = +15\).

   $$
   4 - 6x + 15 = 2x + 2 - 7
   $$
2. Combine like terms on each side.

   $$
   19 - 6x = 2x - 5
   $$
3. Add \(6x\) to both sides and add 5 to both sides.

   $$
   24 = 8x
   $$
4. Divide both sides by 8.

   $$
   x = 3
   $$
5. Check: both sides give 1.

   $$
   4 - 3(2 \cdot 3 - 5) = 1 = 2(3 + 1) - 7
   $$

Answer: \(x = 3\)

**Example 3: fractions with two-term numerators**

Problem: Solve \(\frac{x + 2}{3} - \frac{x - 1}{4} = 2\).

1. The least common denominator is 12. Multiply every term by 12.

   $$
   12 \cdot \frac{x + 2}{3} - 12 \cdot \frac{x - 1}{4} = 12 \cdot 2
   $$
2. Simplify. Keep each numerator in parentheses, because the minus sign belongs to the whole group.

   $$
   4(x + 2) - 3(x - 1) = 24
   $$
3. Distribute. Notice \(-3 \cdot -1 = +3\).

   $$
   4x + 8 - 3x + 3 = 24
   $$
4. Combine like terms.

   $$
   x + 11 = 24
   $$
5. Subtract 11.

   $$
   x = 13
   $$
6. Check: \(\frac{15}{3} - \frac{12}{4} = 5 - 3 = 2\).

   $$
   \frac{13 + 2}{3} - \frac{13 - 1}{4} = 2
   $$

Answer: \(x = 13\)

**Example 4 (test-hard): write the equation yourself**

Problem: A school club sold 200 tickets to a show. Student tickets cost $8 and adult tickets cost $15. The club took in $2,440. How many adult tickets did it sell?

1. Let \(a\) be the number of adult tickets. The other \(200 - a\) tickets were student tickets.
2. Money from adults plus money from students equals the total.

   $$
   15a + 8(200 - a) = 2440
   $$
3. Distribute the 8.

   $$
   15a + 1600 - 8a = 2440
   $$
4. Combine like terms and subtract 1600.

   $$
   7a = 840
   $$
5. Divide by 7.

   $$
   a = 120
   $$
6. Check: 120 adult tickets bring in $1,800 and 80 student tickets bring in $640. Together that is $2,440.

Answer: 120 adult tickets (and 80 student tickets).

## Common mistakes

- **Losing the minus sign in front of parentheses.** \(-3(2x - 5)\) is \(-6x + 15\). Fix: multiply the outside number, sign included, by every term inside.
- **Multiplying only the fractions by the LCD.** In Example 3, the 2 on the right must become 24 too. Fix: every term on both sides gets multiplied.
- **Dropping the parentheses around a numerator.** \(-\frac{x - 1}{4}\) times 12 is \(-3(x - 1)\), which is \(-3x + 3\), not \(-3x - 1\).
- **Dividing too early.** From \(3x - 7 = 11\), dividing by 3 first gives \(x - \frac{7}{3} = \frac{11}{3}\). It still works but invites mistakes. Fix: move the plain numbers first, divide last.
- **Skipping the check.** Substituting takes 20 seconds and catches almost every sign error.

## Quick methods

> **Tip: Clear decimals the same way as fractions**
>
> For \(0.4x + 1.25 = 0.9x - 0.75\), multiply every term by 100 to get \(40x + 125 = 90x - 75\). Whole numbers are much easier to work with, and the answer does not change (here \(x = 4\)).

> **Note: Put the variable on the side where it stays positive**
>
> In Example 2 we moved \(-6x\) to the right, where it joined \(2x\) to make \(8x\). Choosing the side with the bigger x coefficient avoids dividing by a negative.

## Practice

**5 practice questions**

1. Solve \(6x + 4 = 2x - 12\).
   A. \(x = -4\)
   B. \(x = -2\)
   C. \(x = 2\)
   D. \(x = 4\)

   Answer: \(x = -4\). Subtract \(2x\) and subtract 4: \(4x = -16\), so \(x = -4\). Getting \(-2\) means you added \(2x\) instead of subtracting it (\(8x = -16\)). Getting 4 is a dropped sign.

2. Solve \(2(3x - 1) - (x + 4) = 9\).
   A. \(x = \frac{7}{5}\)
   B. \(x = \frac{14}{5}\)
   C. \(x = 3\)
   D. \(x = 5\)

   Answer: \(x = 3\). Distribute: \(6x - 2 - x - 4 = 9\), so \(5x - 6 = 9\) and \(5x = 15\). The answer \(\frac{7}{5}\) comes from writing \(-(x + 4)\) as \(-x + 4\). The answer \(\frac{14}{5}\) comes from multiplying only the \(3x\) by 2.

3. Solve \(\frac{x}{3} + 5 = \frac{x}{2}\).
   A. \(x = -30\)
   B. \(x = 5\)
   C. \(x = 15\)
   D. \(x = 30\)

   Answer: \(x = 30\). Multiply every term by 6: \(2x + 30 = 3x\). Subtract \(2x\): \(x = 30\). Check: \(10 + 5 = 15\), and \(\frac{30}{2} = 15\). If you got 5, you forgot to multiply the 5 by 6. And 15 is the value of each side, not x.

4. Solve \(3x + 2y = 12\) for \(y\).
   A. \(y = 12 - 3x\)
   B. \(y = 6 - 3x\)
   C. \(y = 6 - \frac{3}{2}x\)
   D. \(y = \frac{3}{2}x - 6\)

   Answer: \(y = 6 - \frac{3}{2}x\). Subtract \(3x\): \(2y = 12 - 3x\). Divide **both** terms by 2: \(y = 6 - \frac{3}{2}x\). The choice \(6 - 3x\) divides only the 12 by 2.

5. How many solutions does \(3(x + 2) - x = 2(x + 4)\) have?
   A. No solution
   B. Exactly one, \(x = 0\)
   C. Exactly one, \(x = 1\)
   D. Infinitely many

   Answer: No solution. The left side is \(2x + 6\) and the right side is \(2x + 8\). Subtracting \(2x\) leaves \(6 = 8\), which is false for every x. So there is no solution.

## Frequently asked questions

### What is a multi-step equation?

It is an equation that needs more than two moves to solve, usually because it has parentheses, fractions, or the variable on both sides. The moves are the same ones you use for simple equations. There are just more of them, done in a set order.

### What should I do first when solving an equation?

Clean up each side before moving anything across the equals sign. Distribute, clear fractions, and combine like terms. Once each side is as simple as it can be, move the variable terms to one side and the numbers to the other.

### What does it mean if the variable cancels out?

Look at what is left. A false statement like \(6 = 8\) means no number works, so there is no solution. A true statement like \(5 = 5\) means every number works, so there are infinitely many solutions. It never means the answer is zero.

### How do I check my answer?

Put your value back into the original equation, not one of your later steps, and work out each side separately. If both sides give the same number, you are right. If they differ, look for a sign error in the distributing step first.

## Related

- [How to solve and graph inequalities](https://duckyhelper.com/learn/algebra-1/inequalities/)
- [How to solve absolute value equations and inequalities](https://duckyhelper.com/learn/algebra-1/absolute-value/)
- [Linear equations in one variable on the SAT](https://duckyhelper.com/learn/sat-math/linear-equations/)
- [Algebra 1 study guides](https://duckyhelper.com/learn/algebra-1/)

## Try asking Ducky

- "I keep getting x = 5 on this one but the answer is 3. Can you check my distributing?"
- "Why do I multiply every term by the LCD and not just the fractions?"
- "Give me two harder equations with fractions to try, then check my work."

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