Question: 3. Solve the absolute value inequalities below. Express each solution in interval notation and show the solution on a number line.
(a) |(9 - 2x)/5| + 4 \leq 7
(b) |2x - 13| - 3 > 4
Graph/math position_hint: center \u2014 two separate algebra problems are shown in the middle-left area, with (a) above (b); no plotted graph, only inequality expressions.
User: In simple steps
1. **Problem statement:** Solve the absolute value inequalities and express solutions in interval notation.
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### (a) \( \left| \frac{9 - 2x}{5} \right| + 4 \leq 7 \)
2. **Isolate the absolute value:**
\[ \left| \frac{9 - 2x}{5} \right| \leq 7 - 4 \]
\[ \left| \frac{9 - 2x}{5} \right| \leq 3 \]
3. **Recall the rule:** For \( |A| \leq B \) where \( B \geq 0 \), the solution is \( -B \leq A \leq B \).
4. **Apply the rule:**
\[ -3 \leq \frac{9 - 2x}{5} \leq 3 \]
5. **Multiply all parts by 5 to clear denominator:**
\[ -3 \times 5 \leq 9 - 2x \leq 3 \times 5 \]
\[ -15 \leq 9 - 2x \leq 15 \]
6. **Subtract 9 from all parts:**
\[ -15 - 9 \leq -2x \leq 15 - 9 \]
\[ -24 \leq -2x \leq 6 \]
7. **Divide all parts by -2, remember to reverse inequality signs because dividing by negative:**
\[ \frac{-24}{-2} \geq x \geq \frac{6}{-2} \]
\[ 12 \geq x \geq -3 \]
8. **Rewrite in standard interval notation:**
\[ [-3, 12] \]
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### (b) \( |2x - 13| - 3 > 4 \)
9. **Isolate the absolute value:**
\[ |2x - 13| > 4 + 3 \]
\[ |2x - 13| > 7 \]
10. **Recall the rule:** For \( |A| > B \) where \( B > 0 \), the solution is \( A < -B \) or \( A > B \).
11. **Apply the rule:**
\[ 2x - 13 < -7 \quad \text{or} \quad 2x - 13 > 7 \]
12. **Solve each inequality separately:**
- For \( 2x - 13 < -7 \):
\[ 2x < -7 + 13 \]
\[ 2x < 6 \]
\[ x < 3 \]
- For \( 2x - 13 > 7 \):
\[ 2x > 7 + 13 \]
\[ 2x > 20 \]
\[ x > 10 \]
13. **Write solution in interval notation:**
\[ (-\infty, 3) \cup (10, \infty) \]
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### Final answers:
- (a) \( x \in [-3, 12] \)
- (b) \( x \in (-\infty, 3) \cup (10, \infty) \)