# Two-Step Equations

A two-step equation is an algebraic equation that can be solved in just two steps. The equation is considered solved when the variable is isolated on one side of the equation. Typically, the variable is kept on the left side, while the constants are moved to the right side.

Similar to solving one-step equations, the concept of inverse operations is used here. Keep in mind that addition cancels out subtraction (and vice versa), and multiplication cancels out 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} $$