1. **State the problem:** Solve the equation $$|p-1| - |2-p| = -10$$. 2. **Recall the definition of absolute value:** For any real number $x$, $$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$$. 3. **Identify critical points:** The expressions inside the absolute values change sign at $p=1$ and $p=2$. We will consider three intervals: $(-\infty,1)$, $[1,2]$, and $(2,\infty)$. 4. **Case 1: $p < 1$** - $|p-1| = 1 - p$ because $p-1 < 0$. - $|2-p| = 2 - p$ because $2-p > 0$. Substitute into the equation: $$ (1 - p) - (2 - p) = -10 $$ Simplify: $$ 1 - p - 2 + p = -10 $$ $$ (1 - 2) + (-p + p) = -10 $$ $$ -1 + 0 = -10 $$ $$ -1 = -10 $$ This is false, so no solution in this interval. 5. **Case 2: $1 \leq p \leq 2$** - $|p-1| = p - 1$ because $p-1 \geq 0$. - $|2-p| = 2 - p$ because $2-p \geq 0$. Substitute: $$ (p - 1) - (2 - p) = -10 $$ Simplify: $$ p - 1 - 2 + p = -10 $$ $$ 2p - 3 = -10 $$ Add 3 to both sides: $$ 2p = -10 + 3 $$ $$ 2p = -7 $$ Divide both sides by 2: $$ p = \frac{-7}{2} $$ Check if $p$ is in the interval $[1,2]$: $$ \frac{-7}{2} = -3.5 \notin [1,2] $$ No solution in this interval. 6. **Case 3: $p > 2$** - $|p-1| = p - 1$ because $p-1 > 0$. - $|2-p| = p - 2$ because $2-p < 0$, so absolute value flips sign. Substitute: $$ (p - 1) - (p - 2) = -10 $$ Simplify: $$ p - 1 - p + 2 = -10 $$ $$ (p - p) + (-1 + 2) = -10 $$ $$ 0 + 1 = -10 $$ $$ 1 = -10 $$ False, so no solution in this interval. 7. **Conclusion:** No value of $p$ satisfies the equation $$|p-1| - |2-p| = -10$$. **Final answer:** No solution.