Solve {l}{40+15|x-2|/5=8-3|x-2|?}{text{ro}}

$$\left. \begin{array} { l } { \frac { 40 + 15 | x - 2 | } { 5 } = 8 - 3 | x - 2 | ? } \\ { \text { ro } } \end{array} \right.$$

Solve for x

$$x=2$$

Steps Using Definition of Absolute Value

Multiply both sides of the equation by 5.

$$40+15|x-2|=40-15|x-2|$$

Add 15|x-2| to both sides.

$$40+15|x-2|+15|x-2|=40$$

Combine 15|x-2| and 15|x-2| to get 30|x-2|.

$$40+30|x-2|=40$$

Subtract 40 from both sides.

$$30|x-2|=40-40$$

Subtract 40 from 40 to get 0.

$$30|x-2|=0$$

Divide both sides by 30. Zero divided by any non-zero number gives zero.

$$|x-2|=0$$

Use the definition of absolute value.

$$x-2=0$$

Add 2 to both sides of the equation.

$$x=2$$