To solve the inequality [tex]\(\left|\frac{1}{2} x - 6\right| < \frac{3}{2}\)[/tex], we need to understand that the absolute value inequality [tex]\(|A| < B\)[/tex] can be rewritten as two separate inequalities:
[tex]\[ -B < A < B \][/tex]
Here, [tex]\(A\)[/tex] is [tex]\(\frac{1}{2} x - 6\)[/tex] and [tex]\(B\)[/tex] is [tex]\(\frac{3}{2}\)[/tex]. So, we can rewrite our inequality as:
[tex]\[ -\frac{3}{2} < \frac{1}{2} x - 6 < \frac{3}{2} \][/tex]
Now, we solve these two inequalities separately.
### Solving [tex]\(-\frac{3}{2} < \frac{1}{2} x - 6\)[/tex]
1. Add 6 to both sides to isolate the term with [tex]\(x\)[/tex]:
[tex]\[
-\frac{3}{2} + 6 < \frac{1}{2} x
\][/tex]
2. Convert [tex]\(6\)[/tex] to a fraction with the same denominator:
[tex]\[
-\frac{3}{2} + \frac{12}{2} < \frac{1}{2} x
\][/tex]
3. Simplify the left side:
[tex]\[
\frac{9}{2} < \frac{1}{2} x
\][/tex]
4. Multiply both sides by 2 to solve for [tex]\(x\)[/tex]:
[tex]\[
9 < x \quad \text{or} \quad x > 9
\][/tex]
### Solving [tex]\(\frac{1}{2} x - 6 < \frac{3}{2}\)[/tex]
1. Add 6 to both sides to isolate the term with [tex]\(x\)[/tex]:
[tex]\[
\frac{1}{2} x - 6 + 6 < \frac{3}{2} + 6
\][/tex]
2. Convert [tex]\(6\)[/tex] to a fraction with the same denominator:
[tex]\[
\frac{1}{2} x < \frac{3}{2} + \frac{12}{2}
\][/tex]
3. Simplify the right side:
[tex]\[
\frac{1}{2} x < \frac{15}{2}
\][/tex]
4. Multiply both sides by 2 to solve for [tex]\(x\)[/tex]:
[tex]\[
x < 15
\][/tex]
### Combining the Two Inequalities
Now we combine the results from the two separate inequalities:
[tex]\[
9 < x < 15
\][/tex]
Thus, the solution to the inequality [tex]\(\left|\frac{1}{2} x - 6\right| < \frac{3}{2}\)[/tex] is:
[tex]\[
\boxed{9 < x < 15}
\][/tex]
