1. **State the problem:** Solve the equation $$|2x - 5| + |x + 1| = 11$$ for $x$. 2. **Understand absolute value:** 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. **Find critical points:** The expressions inside the absolute values change sign at points where they equal zero: - $2x - 5 = 0 \Rightarrow x = \frac{5}{2} = 2.5$ - $x + 1 = 0 \Rightarrow x = -1$ These points divide the real line into three intervals: - Interval 1: $x < -1$ - Interval 2: $-1 \leq x < 2.5$ - Interval 3: $x \geq 2.5$ 4. **Solve on each interval:** **Interval 1: $x < -1$** - $2x - 5 < 0$ so $|2x - 5| = -(2x - 5) = -2x + 5$ - $x + 1 < 0$ so $|x + 1| = -(x + 1) = -x - 1$ Equation becomes: $$-2x + 5 - x - 1 = 11$$ $$-3x + 4 = 11$$ $$-3x = 7$$ $$x = -\frac{7}{3} \approx -2.333$$ Check if $x = -\frac{7}{3}$ is in Interval 1 ($x < -1$): Yes, since $-2.333 < -1$. So this solution is valid. **Interval 2: $-1 \leq x < 2.5$** - $2x - 5 < 0$ so $|2x - 5| = -2x + 5$ - $x + 1 \geq 0$ so $|x + 1| = x + 1$ Equation becomes: $$-2x + 5 + x + 1 = 11$$ $$-x + 6 = 11$$ $$-x = 5$$ $$x = -5$$ Check if $x = -5$ is in Interval 2 ($-1 \leq x < 2.5$): No, since $-5 < -1$. So discard this solution. **Interval 3: $x \geq 2.5$** - $2x - 5 \geq 0$ so $|2x - 5| = 2x - 5$ - $x + 1 \geq 0$ so $|x + 1| = x + 1$ Equation becomes: $$2x - 5 + x + 1 = 11$$ $$3x - 4 = 11$$ $$3x = 15$$ $$x = 5$$ Check if $x = 5$ is in Interval 3 ($x \geq 2.5$): Yes, since $5 \geq 2.5$. So this solution is valid. 5. **Final solutions:** $$x = -\frac{7}{3} \quad \text{and} \quad x = 5$$