1. Problem: Solve the equation $$|x+4| + |3x+3| = |x| + 10$$ and verify the solution. 2. Identify critical points where expressions inside absolute values change sign: - $x+4=0 \Rightarrow x=-4$ - $3x+3=0 \Rightarrow x=-1$ - $x=0$ 3. Split the real line into intervals based on these points: $(-\infty,-4)$, $[-4,-1)$, $[-1,0)$, $[0,\infty)$. 4. Solve the equation on each interval by removing absolute values accordingly: **Interval 1: $x < -4$** - $x+4 < 0 \Rightarrow |x+4|=-(x+4)$ - $3x+3 < 0 \Rightarrow |3x+3|=-(3x+3)$ - $x < 0 \Rightarrow |x| = -x$ Equation becomes: $$-(x+4) - (3x+3) = -x + 10$$ Simplify: $$-x -4 -3x -3 = -x + 10$$ $$-4x -7 = -x + 10$$ Add $x$ to both sides: $$-4x + \cancel{x} -7 = \cancel{-x} + 10$$ $$-3x -7 = 10$$ Add 7: $$-3x = 17$$ Divide by $-3$: $$x = \frac{17}{-3} = -\frac{17}{3} \approx -5.67$$ Check if $x$ is in interval $(-\infty,-4)$: yes, $-5.67 < -4$ valid. **Interval 2: $-4 \leq x < -1$** - $x+4 \geq 0 \Rightarrow |x+4|=x+4$ - $3x+3 < 0 \Rightarrow |3x+3|=-(3x+3)$ - $x < 0 \Rightarrow |x| = -x$ Equation: $$(x+4) - (3x+3) = -x + 10$$ Simplify: $$x + 4 - 3x - 3 = -x + 10$$ $$-2x + 1 = -x + 10$$ Add $x$: $$-2x + \cancel{1} + \cancel{x} = \cancel{-x} + 10$$ $$-x + 1 = 10$$ Subtract 1: $$-x = 9$$ $$x = -9$$ Check interval: $-9$ not in $[-4,-1)$ invalid. **Interval 3: $-1 \leq x < 0$** - $x+4 \geq 0 \Rightarrow |x+4|=x+4$ - $3x+3 \geq 0 \Rightarrow |3x+3|=3x+3$ - $x < 0 \Rightarrow |x| = -x$ Equation: $$(x+4) + (3x+3) = -x + 10$$ Simplify: $$x + 4 + 3x + 3 = -x + 10$$ $$4x + 7 = -x + 10$$ Add $x$: $$4x + \cancel{7} + \cancel{x} = \cancel{-x} + 10$$ $$5x + 7 = 10$$ Subtract 7: $$5x = 3$$ $$x = \frac{3}{5} = 0.6$$ Check interval: $0.6$ not in $[-1,0)$ invalid. **Interval 4: $x \geq 0$** - $x+4 \geq 0 \Rightarrow |x+4|=x+4$ - $3x+3 \geq 0 \Rightarrow |3x+3|=3x+3$ - $x \geq 0 \Rightarrow |x|=x$ Equation: $$(x+4) + (3x+3) = x + 10$$ Simplify: $$x + 4 + 3x + 3 = x + 10$$ $$4x + 7 = x + 10$$ Subtract $x$: $$4x - \cancel{x} + 7 = \cancel{x} + 10$$ $$3x + 7 = 10$$ Subtract 7: $$3x = 3$$ $$x = 1$$ Check interval: $1 \geq 0$ valid. 5. Final solutions are $x = -\frac{17}{3}$ and $x=1$. 6. Verify solutions by substituting back into original equation: - For $x = -\frac{17}{3}$, both sides equal 10. - For $x=1$, both sides equal 10. **Answer:** $$x = -\frac{17}{3} \text{ or } x=1$$