Disequazione $|x-3|+|x+2|\le 9$

Text

Solve the two–absolute–value inequality $|x-3|+|x+2|\le 9.$

Solution

The points making the arguments vanish are $x=-2$ and $x=3$; we study three intervals.

• $x<-2$: $-(x-3)-(x+2)=-2x+1\le 9\implies x\ge -4$, so $-4\le x<-2$.
• $-2\le x\le 3$: $(3-x)+(x+2)=5\le 9$, always true, so $-2\le x\le 3$.
• $x>3$: $(x-3)+(x+2)=2x-1\le 9\implies x\le 5$, so $3<x\le 5$. Combining the three cases: $\boxed{-4\le x\le 5.}$