Chapter 1.4, Problem 59E

To determine

Tofind:the solution of the equation gives the reason for the solution.

Answer to Problem 59E

  $x=1,x=-5.$

Explanation of Solution

Given:

  $8\left|x+2\right|-6=5\left|x+2\right|+3$ .

Concept used:

Absolute value or modulus of a real number $x$ ,denoted that $\left|x\right|$ , is the non-negative value of $x$ without regard to its sign.

Mathematically,

  $\left|x\right|=x.$

If domain $x$ is positive.

  $\left|x\right|=-x.$

If domain $x$ is negative.

Calculation:

In solving absolute value equations, remember that $\left|ax+b\right|=c$ provided that $c\ge 0$ will either be:

  $ax+b=c\text{ or }ax+b=-c$ .

Note that when $c<0$ , then the equation has no solution because absolute values are always positive.

A solution is extraneous if it does not satisfy the original equation upon checking.

Let $a=\left|x+2\right|$ .

According to the given:

  $8\left|x+2\right|-6=5\left|x+2\right|+3$ .

Substituting the value as:

  $\begin{array}{l}8a-6=5a+3.\\ a=3.\\ \left|x+2\right|=3.\end{array}$

The absolute value has two value with plus and minus sign.

  $\begin{array}{l}x+2=3\text{ and }x+2=-3.\\ x=1,x=-5.\end{array}$

It can be check by putting the value back to the equations.

By putting $x=1$ :

  $\begin{array}{l}8\left|x+2\right|-6=5\left|x+2\right|+3.\\ 8\left|1+2\right|-6=5\left|1+2\right|+3.\\ 18=18.\end{array}$

By putting $x=-5$ :

  $\begin{array}{l}8\left|x+2\right|-6=5\left|x+2\right|+3.\\ 8\left|-5+2\right|-6=5\left|-5+2\right|+3.\\ 18=18.\end{array}$

Hence, $x=1,x=-5.$