A two-step equation is an algebraic equation that can be solved in just two steps. We know that we have already solved the equation if we are able to isolate the variable by itself on one side of the equation. Most of the time, we keep the variables on the left side and the constants on the right side.

Just like with one-step equations, we will use the idea of inverse operations to solve it. Remember, addition undoes subtraction (and vice versa), and multiplication undoes division (and vice versa).

Example 1: Solve for \(x\).

$$3x + 5 = 11$$

The number \(5\) is being added to \(3x\). The inverse of addition is subtraction. That means we are going to subtract both sides of the equation by \(5\).

$$\eqalign{

3x + 5 &= 11 \cr

3x + 5\color{red}{ - 5} &= 11 \color{red}{- 5} \cr

3x &= 6 \cr} $$

The variable \(x\) is being multiplied by \(3\). To undo it, we will need to divide both sides by \(3\).

$$\eqalign{

{{3x} \over \color{red}{3}} &= {6 \over\color{red}{3}} \cr

x &= 2 \cr} $$

Example 2: Solve for \(y\).

$${y \over 4} - 8 =1$$

The term \(\large{y \over 4}\) is being subtracted by \(8\). The inverse operation of subtracting \(8\) is adding \(8\). Let's add \(8\) to both sides of the equation.

$$\eqalign{

{y \over 4} - 8 &= 1 \cr

{y \over 4} - 8 \color{red}{+ 8} &= 1 \color{red}{+ 8} \cr

{y \over 4} &= 9 \cr} $$

The variable \(y\) is being divided by \(4\). The inverse operation of dividing \(4\) is multiplying \(4\). That means we are going to multiply both sides of the equation by \(4\).

$$\eqalign{

{y \over 4} &= 9 \cr

\color{red}{ 4}\left( {{y \over 4}} \right) &=\color{red}{ 4}\left( 9 \right) \cr

y &= 36 \cr} $$

Example 3: Solve for \(a\).

$${{a + 3} \over 2} = - 4$$

The expression \(a+3\) is being divided by \(2\). The inverse of dividing by \(2\) is multiplying by \(2\). So we first multiply both sides of the equation by \(2\).

$$\eqalign{

{{a + 3} \over 2} &= - 4 \cr

\color{red}{2}\left( {{{a + 3} \over 2}} \right) &= \color{red}{2}\left( { - 4} \right) \cr

a + 3 &= - 8 \cr} $$

Now, \(3\) is being added to \(a\). To undo that, we subtract both sides by \(3\).

$$\eqalign{

a + 3 &= - 8 \cr

a + 3 \color{red}{- 3} &= - 8 \color{red}{- 3} \cr

a &= - 11 \cr} $$