Solving Two-Step Equations

Learning goals

Recognize $ax + b = c$ as two operations on the variable
Undo in reverse order, clearing the constant first and undoing the multiplication or division second
Divide by the full coefficient, sign and all
Combine like terms or distribute first when a side needs cleaning
Leave a non-integer answer as a simplified fraction

What makes an equation a two-step equation

A two-step equation has two operations attached to the variable, so isolating the variable takes two inverse moves instead of one. The most common shape is

ax + b = c,

where $a$ is the coefficient multiplying the variable, $b$ is a constant added on, and $c$ is the number the whole expression equals. In $2x + 3 = 11$ the coefficient is $a = 2$ , the constant is $b = 3$ , and $c = 11$ . Reading the left side the way you would evaluate it, you multiply $x$ by $2$ first, then add $3$ . Two operations went in, so two inverse operations must come out.

Nothing about the balance principle changes. An equation is still a balanced scale, and the two sides hold equal amounts. So a balance-preserving move, adding or subtracting the same number on both sides, or multiplying or dividing both sides by the same nonzero number, keeps the solution unchanged, whichever side of the equation the variable is on. What is new is that you will use that principle twice in a row, and the order in which you use it matters.

Why you undo in the reverse order

To see which operation to undo first, think about how the left side of $2x + 3$ is built up from $x$ . Then you run that construction backward.

Start with the variable $x$ . The order of operations says multiplication happens before addition. So to build $2x + 3$ you first multiply by $2$ , reaching $2x$ , and then add $3$ , reaching $2x + 3$ . The addition is the outermost operation: it is the last thing wrapped around the variable, sitting on the outside of everything else.

To take the expression apart you reverse that process, removing the outermost layer first, the way you cannot reach an onion’s inner skin without peeling the outer one off. So you undo the addition before the multiplication: first subtract $3$ from both sides to strip off the constant, which leaves $2x$ alone. Then you divide by $2$ to strip off the coefficient, which leaves $x$ alone.

This is why the rule is “reverse the order of operations.” Evaluating works inside out, doing multiplication before addition; solving works outside in, undoing addition before multiplication. The two orders are mirror images, and that single fact decides every two-step equation.

The left side of 2x + 3 is built by multiplying x by 2 (inner layer), then adding 3 (outer layer). To isolate x you peel from the outside in: subtract 3 first, then divide by 2.

Why the constant comes off before the coefficient#

Take the standard form $ax + b = c$ with a nonzero coefficient $a$ . Suppose some value of $x$ makes that equation true, so the two sides are the same number.

Subtract $b$ from both sides. The left side becomes $ax + b - b$ , and since $b - b = 0$ , that is just $ax$ ; the right side becomes $c - b$ . Because the same amount was removed from each side, the equation $ax = c - b$ has exactly the same solution as the one you started with. Notice what this step accomplished: the constant $b$ is gone, and the variable term $ax$ now stands by itself. This move could not wait, because while $b$ was still attached by addition, the variable term was not alone on its side. With $b$ still attached, dividing that whole side by $a$ would not have isolated $x$ either.

Now divide both sides of $ax = c - b$ by $a$ . The left side becomes $\dfrac{ax}{a} = x$ , since $\dfrac{a}{a} = 1$ ; the right side becomes $\dfrac{c - b}{a}$ . Again the same operation hit both sides, so the solution is unchanged, and now $x$ stands completely alone:

x = \frac{c - b}{a}.

Clearing the constant first is the clean, efficient order. The constant is the outer layer, added after the multiplication. So undoing that outer layer first removes one layer at a time and leaves the variable term alone. You could divide by $a$ while $b$ is still present, and you would reach the same answer. But that step divides $b$ by $a$ too, and can turn a whole number into a fraction, making the arithmetic messier for no gain. To keep the numbers whole, undo the addition or subtraction first, then the multiplication or division.

The two-step recipe

Every equation in this lesson follows the same two moves once the variable term appears on only one side:

Undo the addition or subtraction. Add or subtract the constant on both sides so the variable term ( $ax$ or $\tfrac{x}{a}$ ) sits by itself.
Undo the multiplication or division. Divide both sides by the coefficient, or multiply both sides by the divisor, to leave the variable alone.

Then check by substituting your answer into the original equation and evaluating with the full order of operations. The check is not optional decoration. With two steps there are two places to slip, and substituting back catches almost every mistake in a single line.

Worked example 1 Solve $2x + 3 = 11$

The variable is multiplied by $2$ and then has $3$ added, so undo the addition first. Subtract $3$ from both sides:

2x + 3 - 3 = 11 - 3.

On the left, $3 - 3 = 0$ , leaving the variable term alone; on the right, $11 - 3 = 8$ :

2x = 8.

Now undo the multiplication by dividing both sides by the coefficient $2$ :

\frac{2x}{2} = \frac{8}{2} \;\Rightarrow\; x = 4.

Check by substituting $4$ into the original equation, doing the multiplication before the addition just as the order of operations requires:

2(4) + 3 = 8 + 3 = 11, \qquad 11 = 11. \;\checkmark

Both sides match, so $x = 4$ . Subtracting the constant first cleared the way to divide off the coefficient.

Worked example 2 Solve $5x - 4 = 16$

This is the subtraction form, $ax - b = c$ . A constant is subtracted from the variable term, so undo that by adding $4$ to both sides:

5x - 4 + 4 = 16 + 4 \;\Rightarrow\; 5x = 20.

The variable term now stands alone. Divide both sides by the coefficient $5$ :

\frac{5x}{5} = \frac{20}{5} \;\Rightarrow\; x = 4.

Check in the original equation, multiplying before subtracting:

5(4) - 4 = 20 - 4 = 16, \qquad 16 = 16. \;\checkmark

So $x = 4$ . A subtracted constant is undone by adding it back, exactly as in the one-step lesson, only now a division step follows.

Check your understanding

To solve $3x + 7 = 22$ , which is the best first step?

A divide both sides by

3

B subtract

7

from both sides C subtract

3

from both sides D add

7

to both sides

Reading the equation in a different order

It is fine, and often clearer, to read the equation in either direction. If the constant comes first, as in $b + ax = c$ , or the whole thing is written backward, as in $c = ax + b$ , the recipe is unchanged. In both of those forms the addition or subtraction is still the outer layer, so undo it first.

Worked example 3 Solve $4 + 3x = 19$

Here the constant $4$ is written before the variable term, but it is still added to $3x$ , so it is still the outer layer. Subtract $4$ from both sides:

4 + 3x - 4 = 19 - 4 \;\Rightarrow\; 3x = 15.

Divide both sides by the coefficient $3$ :

\frac{3x}{3} = \frac{15}{3} \;\Rightarrow\; x = 5.

Check in the original equation:

4 + 3(5) = 4 + 15 = 19, \qquad 19 = 19. \;\checkmark

So $x = 5$ . Writing the constant first does not change anything; you still strip it off before dividing.

Worked example 4 Solve $25 = 7x - 3$

The variable term is on the right, but the balance principle does not care which side the variable lives on. The constant $-3$ is the outer layer on the right, so add $3$ to both sides:

25 + 3 = 7x - 3 + 3 \;\Rightarrow\; 28 = 7x.

Divide both sides by the coefficient $7$ :

\frac{28}{7} = \frac{7x}{7} \;\Rightarrow\; 4 = x.

That is the same as $x = 4$ , since an equation reads the same both ways. Check in the original:

7(4) - 3 = 28 - 3 = 25, \qquad 25 = 25. \;\checkmark

So $x = 4$ . When the variable sits on the right, just work toward isolating it there; you do not need to swap the sides first.

Negatives, fractions, and decimals ride along

The recipe does not flinch when the numbers get awkward. A negative coefficient, a fractional answer, or a decimal constant changes the arithmetic of each step but not the two steps themselves. You still clear the constant, then divide by the coefficient, sign and all. For instance, $2x + 1.5 = 5.5$ subtracts to $2x = 4$ and divides to $x = 2$ , the same two moves with decimal arithmetic in place of whole numbers.

Worked example 5 Solve $-3x + 8 = 2$

The constant $8$ is added on, so subtract $8$ from both sides first:

-3x + 8 - 8 = 2 - 8 \;\Rightarrow\; -3x = -6.

Now divide both sides by the full coefficient $-3$ , carrying its sign. A negative divided by a negative is positive, the rule from the integers chapter:

\frac{-3x}{-3} = \frac{-6}{-3} \;\Rightarrow\; x = 2.

Check in the original equation:

-3(2) + 8 = -6 + 8 = 2, \qquad 2 = 2. \;\checkmark

So $x = 2$ . The key is dividing by $-3$ , not by a bare $3$ , so the sign of the answer comes out right.

Check your understanding

After clearing the constant in $-4x + 9 = 1$ you reach $-4x = -8$ . What is $x$ ?

x = 2

x = -2

x = 32

x = -32

Worked example 6 Solve $4x + 1 = 7$

Subtract the constant $1$ from both sides:

4x + 1 - 1 = 7 - 1 \;\Rightarrow\; 4x = 6.

Divide both sides by $4$ . The quotient is not a whole number, which is perfectly fine; write it as a fraction in lowest terms:

x = \frac{6}{4} = \frac{3}{2}.

Check by substituting $\tfrac{3}{2}$ into the original equation:

4 \cdot \tfrac{3}{2} + 1 = \tfrac{12}{2} + 1 = 6 + 1 = 7, \qquad 7 = 7. \;\checkmark

So $x = \tfrac{3}{2}$ . A non-integer answer is not a sign of a mistake; simplify the fraction and check it the same way.

Check your understanding

Solve $2x - 5 = 8$ for $x$ .

x = \tfrac{13}{2}

x = \tfrac{3}{2}

x = \tfrac{3}{4}

x = 13

The divide form: a variable split by a number

When the variable is divided by a number and then has a constant added, the equation looks like

\frac{x}{a} + b = c.

The two layers are now “divide by $a$ ” (inner) and “add $b$ ” (outer), so the recipe runs the same way. That means you clear the constant first, then undo the division by multiplying both sides by $a$ .

Worked example 7 Solve $\dfrac{x}{3} + 2 = 9$

The constant $2$ is the outer layer, so subtract it from both sides:

\frac{x}{3} + 2 - 2 = 9 - 2 \;\Rightarrow\; \frac{x}{3} = 7.

The variable is now divided by $3$ and nothing else, so undo that by multiplying both sides by $3$ :

3 \cdot \frac{x}{3} = 3 \cdot 7 \;\Rightarrow\; x = 21.

Check in the original equation, dividing before adding:

\frac{21}{3} + 2 = 7 + 2 = 9, \qquad 9 = 9. \;\checkmark

So $x = 21$ . The only change from the multiplication form is the second step: a divided variable is freed by multiplying, not dividing.

Check your understanding

After clearing the constant in $\dfrac{x}{4} - 1 = 5$ you reach $\dfrac{x}{4} = 6$ . What is the correct second step?

A divide both sides by

4

B multiply both sides by

4

C subtract

4

from both sides D multiply both sides by

6

When a side needs cleaning up first

Some equations are not yet in two-step form, but a single move from earlier in this chapter turns them into one. Before you reach for an inverse operation, look at each side and simplify it first.

If one side has like terms, combine them. Recall from the simplifying lesson that $3x + 2x = 5x$ , because the two terms count the same thing. So $3x + 2x + 1 = 11$ becomes $5x + 1 = 11$ , an ordinary two-step equation.

Worked example 8 Solve $3x + 2x + 1 = 11$

The left side has two like terms, $3x$ and $2x$ . Combine them before doing anything else:

3x + 2x + 1 = 11 \;\Rightarrow\; 5x + 1 = 11.

Now it is a standard two-step equation. Subtract the constant $1$ from both sides:

5x + 1 - 1 = 11 - 1 \;\Rightarrow\; 5x = 10.

Divide both sides by the coefficient $5$ :

\frac{5x}{5} = \frac{10}{5} \;\Rightarrow\; x = 2.

Check in the original equation, before any simplifying, so the check tests your whole solution:

3(2) + 2(2) + 1 = 6 + 4 + 1 = 11, \qquad 11 = 11. \;\checkmark

So $x = 2$ . Combining like terms collapsed three terms into the familiar $ax + b = c$ .

If a side is a number times a sum, distribute first. The distributive property from the start of this chapter says $2(x + 3) = 2x + 6$ , multiplying the $2$ across both terms inside the parentheses. That rewrite clears the parentheses and again leaves a two-step equation. (When a single number multiplies an entire side and nothing else is added outside the parentheses, dividing both sides by that number first works too. Distributing is the move you will lean on once a side mixes parentheses with other terms.)

Worked example 9 Solve $2(x + 3) = 14$

The left side is $2$ times the sum $x + 3$ . Distribute the $2$ across both terms inside the parentheses:

2(x + 3) = 14 \;\Rightarrow\; 2x + 6 = 14.

Now solve the two-step equation. Subtract the constant $6$ from both sides:

2x + 6 - 6 = 14 - 6 \;\Rightarrow\; 2x = 8.

Divide both sides by the coefficient $2$ :

\frac{2x}{2} = \frac{8}{2} \;\Rightarrow\; x = 4.

Check in the original equation, evaluating the parentheses first as the order of operations demands:

2(4 + 3) = 2(7) = 14, \qquad 14 = 14. \;\checkmark

So $x = 4$ . Distributing cleared the parentheses and turned the equation into one you already know how to finish.

Check your understanding

Which is the best first step toward solving $3(x + 2) = 21$ ?

A distribute the

3

across

x

and

2

B subtract

2

from both sides C divide both sides by

x

D add

3

to both sides

Check your understanding

Which is the best first step toward solving $4x - x + 2 = 11$ ?

A subtract

2

from both sides B combine

4x - x

into

3x

C divide both sides by

4

D combine

4x - x

into

4

Reading a two-step equation before you move

Every two-step equation comes down to the same short routine. First, simplify each side if it has like terms or parentheses. Then ask what is the outer operation (the addition or subtraction) and undo it on both sides. Finally ask what is the inner operation (the multiplication or division) and undo that on both sides. The table below is the whole lesson at a glance.

If the equation reads	step 1: undo the constant	step 2: undo the coefficient	which gives
$ax + b = c$	subtract $b$	divide by $a$	$x = \dfrac{c - b}{a}$
$ax - b = c$	add $b$	divide by $a$	$x = \dfrac{c + b}{a}$
$\dfrac{x}{a} + b = c$	subtract $b$	multiply by $a$	$x = a(c - b)$
$\dfrac{x}{a} - b = c$	add $b$	multiply by $a$	$x = a(c + b)$

The pattern to lock in is the order: the outer layer, the added or subtracted constant, comes off first. The inner layer, the coefficient or divisor, comes off second, and you finish by substituting your answer back to confirm both sides are equal.

Worked example 10 Identify the moves, then solve $\dfrac{x}{2} - 4 = -1$

First read the equation. The variable is divided by $2$ (inner) and then has $4$ subtracted (outer), so this is the divide form. Undo the subtraction first by adding $4$ to both sides:

\frac{x}{2} - 4 + 4 = -1 + 4 \;\Rightarrow\; \frac{x}{2} = 3.

Now undo the division by multiplying both sides by $2$ :

2 \cdot \frac{x}{2} = 2 \cdot 3 \;\Rightarrow\; x = 6.

Check in the original equation:

\frac{6}{2} - 4 = 3 - 4 = -1, \qquad -1 = -1. \;\checkmark

So $x = 6$ . Reading the form first told you to add before multiplying, and the check confirms it.

Common mistakes

Subtracting the coefficient instead of dividing. Once you reach $2x = 8$ , the $2$ is multiplied by the variable, so you divide both sides by $2$ . Subtracting $2$ does nothing useful, because $2x$ is not $2 + x$ .
Forgetting to do both steps to both sides. Each step is a balance move: subtract the constant from both sides, then divide both sides by the coefficient. Touch only one side and the scale tips.
Combining unlike terms. In $5x + 1 = 11$ the $5x$ and the $1$ are not like terms, so they cannot be merged into $6x$ or $6$ . Only $5x$ and another $x$ -term combine; a number combines only with another number.
Distributing to only one term. $2(x + 3)$ is $2x + 6$ , not $2x + 3$ . The factor multiplies every term inside the parentheses, not just the first.
Losing a negative sign. When the coefficient is $-3$ , divide by $-3$ , sign included, so the answer has the right sign. Dividing $-6$ by a bare $3$ drops the negative and flips the answer.
Skipping the check. With two steps there are two chances to slip. Substituting your answer back into the original equation, with the full order of operations, costs one line and catches almost every error.

Key takeaways

A two-step equation has two operations attached to the variable, most often the form $ax + b = c$ , where $a$ is the coefficient and $b$ is the constant.
Undo the operations in the reverse of the order of operations. The added or subtracted constant is the outer layer, so clear it first; the coefficient or divisor is the inner layer, so undo it second.
The recipe: first add or subtract to clear the constant, isolating the variable term. Then divide by the coefficient or multiply by the divisor to isolate the variable.
Negative, fraction, and decimal numbers change the arithmetic of each step, not the steps. Divide by the full coefficient, sign and all, and leave a non-integer answer as a simplified fraction.
If a side has like terms, combine them first; if a side is a number times a sum, distribute first. Either move turns a longer equation back into a two-step.
Always check by substituting your answer into the original equation and evaluating with the full order of operations.

Solving Two-Step Equations

Learning goals

What makes an equation a two-step equation

Why you undo in the reverse order

The two-step recipe

Reading the equation in a different order

Negatives, fractions, and decimals ride along

The divide form: a variable split by a number

When a side needs cleaning up first

Reading a two-step equation before you move

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice