Sum of two absolute values less than or equal to 6 (|x-1|+|x+2| ≤ 6)

Statement

Solve the inequality $|x-1|+|x+2|\le6$.

Solution

There is a sum of two absolute values: we use the sign table. The arguments change sign at $x=1$ and $x=-2$.

• $x<-2$: $-(x-1)-(x+2)\le6 \iff -2x-1\le6 \iff x\ge-\frac72$, that is $-\frac72\le x<-2$;
• $-2\le x<1$: $-(x-1)+(x+2)=3\le6$, always true, the whole of $[-2,1)$;
• $x\ge1$: $(x-1)+(x+2)\le6 \iff 2x+1\le6 \iff x\le\frac52$, that is $1\le x\le\frac52$.

Union: $\boxed{-\dfrac72\le x\le\dfrac52}$.

Geometrically: $|x-1|+|x+2|$ is the sum of the distances of $x$ from the points $1$ and $-2$; since their distance is $3$, the sum is at most $6$ as long as $x$ stays at a distance $\frac{6-3}{2}$ from the interval $[-2,1]$.

Links

Topics: Parabola
Concepts: Absolute value inequalities · Sign table · Absolute value
Skills: Reasoning by cases · Solving inequalities
Exercise type: Solving an inequality