1. **State the problem:** Solve the equation involving absolute values: $|2 - x| + |3 + x| = 5$. 2. **Understand absolute values:** The absolute value $|a|$ equals $a$ if $a \geq 0$, and $-a$ if $a < 0$. We must consider cases based on the expressions inside the absolute values. 3. **Identify critical points:** The expressions inside the absolute values change sign at points where $2 - x = 0$ and $3 + x = 0$. - For $2 - x = 0$, $x = 2$. - For $3 + x = 0$, $x = -3$. 4. **Split into intervals:** We consider three intervals based on these points: - Interval 1: $x < -3$ - Interval 2: $-3 \leq x \leq 2$ - Interval 3: $x > 2$ 5. **Solve in each interval:** **Interval 1: $x < -3$** - $2 - x > 0$ because $x < -3$ implies $2 - x > 0$ (since $2 - (-4) = 6 > 0$) - $3 + x < 0$ because $x < -3$ So, $$|2 - x| = 2 - x$$ $$|3 + x| = -(3 + x) = -3 - x$$ Equation becomes: $$2 - x - 3 - x = 5$$ $$-1 - 2x = 5$$ $$-2x = 6$$ $$x = -3$$ Check if $x = -3$ is in interval 1: no, it is the boundary. **Interval 2: $-3 \leq x \leq 2$** - $2 - x \geq 0$ so $|2 - x| = 2 - x$ - $3 + x \geq 0$ so $|3 + x| = 3 + x$ Equation: $$2 - x + 3 + x = 5$$ $$5 = 5$$ This is true for all $x$ in $[-3, 2]$. **Interval 3: $x > 2$** - $2 - x < 0$ so $|2 - x| = -(2 - x) = x - 2$ - $3 + x > 0$ so $|3 + x| = 3 + x$ Equation: $$x - 2 + 3 + x = 5$$ $$2x + 1 = 5$$ $$2x = 4$$ $$x = 2$$ Check if $x = 2$ is in interval 3: no, it is the boundary. 6. **Combine solutions:** - From interval 2, all $x$ in $[-3, 2]$ satisfy the equation. - At boundaries $x = -3$ and $x = 2$, the equation holds. **Final solution:** $$\boxed{[-3, 2]}$$