Absolute Value Equations - College Algebra

1. The solution set of $||x| - 3| = 2$ is:
   Answer: $\{-5, -1, 1, 5\}$ For $||x| - 3| = 2$, we first solve $|x| - 3 = 2$ or $|x| - 3 = -2$. From $|x| - 3 = 2$: $|x| = 5$, so $x = 5$ or $x = -5$. From $|x| - 3 = -2$: $|x| = 1$, so $x = 1$ or $x = -1$. Let's verify: For $x = 5$: $||5| - 3| = |5 - 3| = |2| = 2$ ✓. For $x = -5$: $||-5| - 3| = |5 - 3| = |2| = 2$ ✓. For $x = 1$: $||1| - 3| = |1 - 3| = |-2| = 2$ ✓. For $x = -1$: $||-1| - 3| = |1 - 3| = |-2| = 2$ ✓. All four solutions are valid.
2. The equation $|3x - 1| = |2x + 5|$ has solution(s):
   Answer: $x = -\frac{4}{5}$ and $x = 6$ For $|3x - 1| = |2x + 5|$, we have two cases: Case 1: $3x - 1 = 2x + 5$, which gives $x = 6$. Case 2: $3x - 1 = -(2x + 5) = -2x - 5$, so $3x - 1 = -2x - 5$, giving $5x = -4$, so $x = -\frac{4}{5}$. Let's verify: For $x = 6$: $|3(6) - 1| = |17| = 17$ and $|2(6) + 5| = |17| = 17$ ✓. For $x = -\frac{4}{5}$: $|3(-\frac{4}{5}) - 1| = |-\frac{12}{5} - 1| = |-\frac{17}{5}| = \frac{17}{5}$ and $|2(-\frac{4}{5}) + 5| = |-\frac{8}{5} + 5| = |\frac{17}{5}| = \frac{17}{5}$ ✓. Both solutions are valid.
3. If $|x - 2|^2 = 9$, then the product of all solutions is:
   Answer: $-5$ From $|x - 2|^2 = 9$, we get $|x - 2| = 3$ (since absolute value is non-negative). This gives us two equations: $x - 2 = 3$ or $x - 2 = -3$. Solving: $x = 5$ or $x = -1$. The product of all solutions is $5 \times (-1) = -5$.
4. The number of solutions to $|x + 3| = 2x - 5$ is:
   Answer: 1 For $|x + 3| = 2x - 5$ to have solutions, we need $2x - 5 \geq 0$, so $x \geq \frac{5}{2}$. Case 1: If $x + 3 \geq 0$ (i.e., $x \geq -3$), then $x + 3 = 2x - 5$, giving $x = 8$. Since $8 \geq \frac{5}{2}$ and $8 \geq -3$, this solution is valid. Case 2: If $x + 3 < 0$ (i.e., $x < -3$), then $-(x + 3) = 2x - 5$, so $-x - 3 = 2x - 5$, giving $3x = 2$, so $x = \frac{2}{3}$. But $\frac{2}{3} \not< -3$, so this contradicts our case assumption. Therefore, there is exactly 1 solution: $x = 8$.

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