1. **Problem (a): Sketch the graph of** $y = |2x - 3|$. 2. The absolute value function $y = |u|$ outputs the distance of $u$ from zero, so it is always non-negative. 3. For $y = |2x - 3|$, the expression inside the absolute value is $2x - 3$. 4. The graph will be a V-shape with the vertex where $2x - 3 = 0$, i.e., at $x = \frac{3}{2}$. 5. For $x \geq \frac{3}{2}$, $2x - 3 \geq 0$, so $y = 2x - 3$ (a line with slope 2). 6. For $x < \frac{3}{2}$, $2x - 3 < 0$, so $y = -(2x - 3) = -2x + 3$ (a line with slope -2). 7. **Problem (b): Solve the inequality** $|2x - 3| < 3x + 2$. 8. First, note that $3x + 2$ must be positive for the inequality to hold, so solve $3x + 2 > 0 \Rightarrow x > -\frac{2}{3}$. 9. For $x > -\frac{2}{3}$, split into two cases: - Case 1: $2x - 3 \geq 0 \Rightarrow x \geq \frac{3}{2}$, then $|2x - 3| = 2x - 3$. Inequality: $2x - 3 < 3x + 2 \Rightarrow -3 < x + 2 \Rightarrow x > -5$. Since $x \geq \frac{3}{2}$ and $x > -5$, solution here is $x \geq \frac{3}{2}$. - Case 2: $2x - 3 < 0 \Rightarrow x < \frac{3}{2}$, then $|2x - 3| = -(2x - 3) = -2x + 3$. Inequality: $-2x + 3 < 3x + 2 \Rightarrow 3 < 5x + 2 \Rightarrow 1 < 5x \Rightarrow x > \frac{1}{5}$. Also, $x > -\frac{2}{3}$ from step 8, so solution here is $\frac{1}{5} < x < \frac{3}{2}$. 10. Combine all: $x > -\frac{2}{3}$ and $x > \frac{1}{5}$, so overall solution is $x > \frac{1}{5}$. 11. **Problem: Solve the inequality** $|5x - 3| < 2|3x - 7|$. 12. Consider cases based on signs of $5x - 3$ and $3x - 7$. 13. Critical points: $x = \frac{3}{5}$ and $x = \frac{7}{3}$. 14. Case 1: $x \geq \frac{7}{3}$, both expressions positive. Inequality: $5x - 3 < 2(3x - 7) \Rightarrow 5x - 3 < 6x - 14 \Rightarrow -3 + 14 < 6x - 5x \Rightarrow 11 < x \Rightarrow x > 11$. Since $x \geq \frac{7}{3}$, solution here is $x > 11$. 15. Case 2: $\frac{3}{5} \leq x < \frac{7}{3}$, $5x - 3 \geq 0$, $3x - 7 < 0$. Inequality: $5x - 3 < 2(7 - 3x) \Rightarrow 5x - 3 < 14 - 6x \Rightarrow 5x + 6x < 14 + 3 \Rightarrow 11x < 17 \Rightarrow x < \frac{17}{11}$. Combine with domain: $\frac{3}{5} \leq x < \frac{7}{3}$ and $x < \frac{17}{11}$, so $\frac{3}{5} \leq x < \frac{17}{11}$. 16. Case 3: $x < \frac{3}{5}$, both expressions negative. Inequality: $-(5x - 3) < 2(7 - 3x) \Rightarrow -5x + 3 < 14 - 6x \Rightarrow -5x + 6x < 14 - 3 \Rightarrow x < 11$. Since $x < \frac{3}{5}$, solution here is $x < \frac{3}{5}$. 17. Combine all: $x < \frac{3}{5}$ or $\frac{3}{5} \leq x < \frac{17}{11}$ or $x > 11$. 18. **Problem: Solve the inequality** $|2x - 1| > 3|x + 2|$. 19. Critical points: $x = \frac{1}{2}$ and $x = -2$. 20. Case 1: $x \geq \frac{1}{2}$, both positive. Inequality: $2x - 1 > 3(x + 2) \Rightarrow 2x - 1 > 3x + 6 \Rightarrow -1 - 6 > 3x - 2x \Rightarrow -7 > x \Rightarrow x < -7$. Contradicts $x \geq \frac{1}{2}$, no solution here. 21. Case 2: $-2 \leq x < \frac{1}{2}$, $2x - 1 < 0$, $x + 2 \geq 0$. Inequality: $-(2x - 1) > 3(x + 2) \Rightarrow -2x + 1 > 3x + 6 \Rightarrow 1 - 6 > 3x + 2x \Rightarrow -5 > 5x \Rightarrow x < -1$. Combine with domain: $-2 \leq x < \frac{1}{2}$ and $x < -1$, so $-2 \leq x < -1$. 22. Case 3: $x < -2$, both negative. Inequality: $-(2x - 1) > -3(x + 2) \Rightarrow -2x + 1 > -3x - 6 \Rightarrow -2x + 3x > -6 - 1 \Rightarrow x > -7$. Combine with domain: $x < -2$ and $x > -7$, so $-7 < x < -2$. 23. Combine all: $-7 < x < -2$ or $-2 \leq x < -1$. 24. **Problem: Solve the inequality** $|3x - a| > 2|x + 2a|$, where $a > 0$. 25. Critical points: $x = \frac{a}{3}$ and $x = -2a$. 26. Case 1: $x \geq \frac{a}{3}$, both positive. Inequality: $3x - a > 2(x + 2a) \Rightarrow 3x - a > 2x + 4a \Rightarrow 3x - 2x > 4a + a \Rightarrow x > 5a$. Combine with domain: $x \geq \frac{a}{3}$ and $x > 5a$, so $x > 5a$. 27. Case 2: $-2a \leq x < \frac{a}{3}$, $3x - a < 0$, $x + 2a \geq 0$. Inequality: $-(3x - a) > 2(x + 2a) \Rightarrow -3x + a > 2x + 4a \Rightarrow a - 4a > 2x + 3x \Rightarrow -3a > 5x \Rightarrow x < -\frac{3a}{5}$. Combine with domain: $-2a \leq x < \frac{a}{3}$ and $x < -\frac{3a}{5}$, so $-2a \leq x < -\frac{3a}{5}$. 28. Case 3: $x < -2a$, both negative. Inequality: $-(3x - a) > -2(x + 2a) \Rightarrow -3x + a > -2x - 4a \Rightarrow -3x + 2x > -4a - a \Rightarrow -x > -5a \Rightarrow x < 5a$. Combine with domain: $x < -2a$ and $x < 5a$, so $x < -2a$. 29. Combine all: $x < -2a$ or $-2a \leq x < -\frac{3a}{5}$ or $x > 5a$. **Final answers:** - (a) Graph is V-shaped with vertex at $x=\frac{3}{2}$. - (b) $x > \frac{1}{5}$. - $|5x - 3| < 2|3x - 7|$ solution: $x < \frac{3}{5}$ or $\frac{3}{5} \leq x < \frac{17}{11}$ or $x > 11$. - $|2x - 1| > 3|x + 2|$ solution: $-7 < x < -2$ or $-2 \leq x < -1$. - $|3x - a| > 2|x + 2a|$ solution: $x < -2a$ or $-2a \leq x < -\frac{3a}{5}$ or $x > 5a$.