To solve the equation [tex]\( \left| \frac{1}{2} x + 5 \right| = \left| -2x - 7 \right| \)[/tex], we need to consider the properties of absolute value functions. An equation involving absolute values, like this one, can be split into multiple cases where the expressions inside the absolute values are either both positive or both negative. Specifically, we handle each part by considering different possible conditions.
Let's solve the equation step by step:
### Case 1: [tex]\(\frac{1}{2} x + 5 \)[/tex] and [tex]\( -2x - 7 \)[/tex] have the same sign.
1) [tex]\( \frac{1}{2} x + 5 = -2x - 7 \)[/tex]
[tex]\[
\frac{1}{2} x + 2x = -7 - 5
\][/tex]
[tex]\[
\frac{1}{2} x + 2x = -12
\][/tex]
Multiply by 2 to eliminate fraction:
[tex]\[
x + 4x = -24
\][/tex]
[tex]\[
5x = -24
\][/tex]
[tex]\[
x = \frac{-24}{5}
\][/tex]
2) [tex]\( \frac{1}{2} x + 5 = 2x + 7 \)[/tex]
[tex]\[
\frac{1}{2} x - 2x = 7 - 5
\][/tex]
[tex]\[
\frac{1}{2} x - 2x = 2
\][/tex]
[tex]\[
-\frac{3}{2} x = 2
\][/tex]
[tex]\[
x = -\frac{2}{\frac{3}{2}}
\][/tex]
[tex]\[
x = -\frac{4}{3}
\][/tex]
### Case 2: [tex]\(\frac{1}{2} x + 5 \)[/tex] and [tex]\( -2x - 7 \)[/tex] have opposite signs.
1) [tex]\( \frac{1}{2} x + 5 = 2x + 7 \)[/tex]
[tex]\[
\frac{1}{2} x - 2x = 7 - 5
\][/tex]
[tex]\[
-\frac{3}{2} x = 2
\][/tex]
[tex]\[
x = -\frac{4}{3}
\][/tex]
2) [tex]\( \frac{1}{2} x + 5 = -2x - 7 \)[/tex]
[tex]\[
\frac{1}{2} x + 2x = - 7 - 5
\][/tex]
[tex]\[
\frac{x}{2} + 2x = -12
\][/tex]
[tex]\[
x/2 + 4x = - 12
\][/tex]
[tex]\[
5/2 x = - 12
\][/tex]
[tex]\[
5x = -24
\][/tex]
[tex]\[
x = -24/5
\][/tex]
### Verification:
It is always good to check both solutions whether they fit into the condition:
Both solutions, [tex]\( x = -\frac{24}{5} \)[/tex] and [tex]\( x = -\frac{4}{3} \)[/tex] satisfy the original equation [tex]\( \left| \frac{1}{2} x + 5 \right| = \left| -2x - 7 \right| \)[/tex].
Thus, the correct answer is:
B. [tex]\( \left\{ \frac{-24}{5}, \frac{-4}{3} \right\} \)[/tex]
