College

Solve [tex]4|x+7|+8=32[/tex]

A. [tex]x=1[/tex] and [tex]x=-13[/tex]

B. [tex]x=-1[/tex] and [tex]x=-13[/tex]

C. [tex]x=-1[/tex] and [tex]x=13[/tex]

D. [tex]x=1[/tex] and [tex]x=-1[/tex]

Answer :

We start with the equation
[tex]$$
4|x+7|+8=32.
$$[/tex]

Step 1: Isolate the Absolute Value

Subtract [tex]$8$[/tex] from both sides:
[tex]$$
4|x+7| = 32 - 8 = 24.
$$[/tex]

Step 2: Divide by [tex]$4$[/tex]

Divide both sides by [tex]$4$[/tex] to isolate the absolute value term:
[tex]$$
|x+7| = \frac{24}{4} = 6.
$$[/tex]

Step 3: Solve the Absolute Value Equation

The equation [tex]$|x+7|=6$[/tex] leads to two cases:

1. Case 1:
[tex]$$
x+7 = 6 \quad \Rightarrow \quad x = 6 - 7 = -1.
$$[/tex]

2. Case 2:
[tex]$$
x+7 = -6 \quad \Rightarrow \quad x = -6 - 7 = -13.
$$[/tex]

Thus, the solutions are [tex]$x=-1$[/tex] and [tex]$x=-13$[/tex].

The correct answer is:

B. [tex]$x=-1$[/tex] and [tex]$x=-13$[/tex].