Absolute value of x-2 less than or equal to x+1 (|x-2| ≤ x+1)

Statement

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

Solution

It is of the type $|A|\le B$: it is equivalent to the system $\begin{cases} x+1\ge0 \\ -(x+1)\le x-2\le x+1 \end{cases}$.

• Condition $x+1\ge0 \iff x\ge-1$;
• Right part: $x-2\le x+1 \iff -2\le1$, always true;
• Left part: $-(x+1)\le x-2 \iff 1\le2x \iff x\ge\frac12$.

Intersecting: $\boxed{x\ge\dfrac12}$.

Links

Topics: Parabola
Concepts: Absolute value inequalities · Absolute value
Skills: Solving inequalities · Solving systems
Exercise type: Solving an inequality