Sure, let's solve the equation [tex]\(\left|\frac{1}{2} x + 3\right| = x + 3\)[/tex].
An absolute value equation [tex]\(|A| = B\)[/tex] holds true if and only if [tex]\(A = B\)[/tex] or [tex]\(A = -B\)[/tex]. Therefore, we split the given equation into two cases:
1. [tex]\(\frac{1}{2} x + 3 = x + 3\)[/tex]
2. [tex]\(\frac{1}{2} x + 3 = -(x + 3)\)[/tex]
Let's solve each case separately:
### Case 1: [tex]\(\frac{1}{2} x + 3 = x + 3\)[/tex]
Subtract 3 from both sides:
[tex]\[
\frac{1}{2} x = x
\][/tex]
Subtract [tex]\(\frac{1}{2} x\)[/tex] from both sides:
[tex]\[
0 = \frac{1}{2} x
\][/tex]
Divide both sides by [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[
x = 0
\][/tex]
Now, verify the solution [tex]\(x = 0\)[/tex] by substituting it back into the original equation:
[tex]\[
\left|\frac{1}{2}(0) + 3\right| = 0 + 3
\][/tex]
Simplifying both sides:
[tex]\[
|3| = 3
\][/tex]
This is a true statement, so [tex]\(x = 0\)[/tex] is a valid solution.
### Case 2: [tex]\(\frac{1}{2} x + 3 = -(x + 3)\)[/tex]
Distribute the negative sign on the right-hand side:
[tex]\[
\frac{1}{2} x + 3 = -x - 3
\][/tex]
Add [tex]\(x\)[/tex] to both sides:
[tex]\[
\frac{3}{2} x + 3 = -3
\][/tex]
Subtract 3 from both sides:
[tex]\[
\frac{3}{2} x = -6
\][/tex]
Divide both sides by [tex]\(\frac{3}{2}\)[/tex]:
[tex]\[
x = -6 \div \frac{3}{2} = -6 \times \frac{2}{3} = -4
\][/tex]
Now, verify the solution [tex]\(x = -4\)[/tex] by substituting it back into the original equation:
[tex]\[
\left|\frac{1}{2}(-4) + 3\right| = -4 + 3
\][/tex]
Simplifying both sides:
[tex]\[
|-2 + 3| = -1
\][/tex]
[tex]\[
|1| = -1
\][/tex]
This is not a true statement, hence [tex]\(x = -4\)[/tex] is not a valid solution.
### Conclusion
The only solution to the equation [tex]\(\left|\frac{1}{2} x + 3\right| = x + 3\)[/tex] is:
[tex]\[
\boxed{0}
\][/tex]
