When we solve an equation, we look for the variable value that will make the equation true. We call this value the solution to the equation. We know that the equation is completely solved if the variable that we are trying to find is isolated (by itself) on one side of the equation. For uniformity, we tend to keep the variables on the left side of the equation.

One-step equations are simple equations that can be solved in a single step. As you will see in the following examples, you can solve it by adding, subtracting, multiplying, or dividing some number. We will use the concept of inverse operation to isolate the variable in question. Simply put, inverse operations are operations that are opposite each other. For example, the addition operation would undo what the subtraction operation does, and vice versa. In the same manner, the multiplication operation would undo what the division operation does, and vice versa.

**Operation \(\Rightarrow\) Inverse Operation**

Addition \( \Rightarrow\) Subtraction

Subtraction \( \Rightarrow\) Addition

Multiplication \( \Rightarrow\) Division

Division \( \Rightarrow\) Multiplication

Example 1: Solve for \(x\).

$$x - 2 = 8$$

Observe that \(2\) is being subtracted from \(x\). The inverse of subtracting \(2\) is adding \(2\). So that's what we are going to do. Add \(2\) to both sides of the equation.

$$\eqalign{

x - 2 &= 8 \cr

x - 2\color{red}{ + 2} &= 8 \color{red}{+ 2} \cr

x &= 10 \cr} $$

Example 2: Solve for \(y\).

$$y + 5 = 12$$

Notice that \(5\) is being added to \(y\). The inverse of adding \(5\) is subtracting \(5\). So we will subtract \(5\) to both sides of the equation.

$$\eqalign{

y + 5 &= 12 \cr

y + 5 \color{red}{- 5} &= 12 \color{red}{- 5} \cr

y &= 7 \cr} $$

Example 3: Solve for \(k\).

$$3k = 18$$

How is \(3\) attached to the variable \(k\)? Clearly, \(k\) is being multiplied by \(3\). The inverse of multiplying by \(3\) is dividing by \(3\). Therefore, we will divide both sides of the equation by \(3\)

$$\eqalign{

3k &= 18 \cr

{{3k} \over \color{red}{3}} &= {{18} \over\color{red}{ 3}} \cr

k &= 6 \cr} $$

Example 4: Solve for \(b\)

$${b \over 7} = 4$$

Observe that \(b\) is being divided by \(7\). The inverse of dividing by \(7\) is multiplying by \(7\). That means we are going to multiply both sides of the equation by \(7\).

$$\eqalign{

{b \over 7} &= 4 \cr

\color{red}{ 7}\left( {{b \over 7}} \right) &= \color{red}{7}\left( 4 \right) \cr

b &= 28 \cr} $$