Equations and inequalities with absolute values

Text

Solve:

• (a) $2|x+3|-x+1=8$
• (b) $|2-x|-x+|x+1|-3=0$
• (c) $|3+x|+4\ge 0$
• (d) $x|x-1|-3x\ge -3$

Solution

(a) $2|x+3|=7+x$ (needs $x\ge-7$). If $x\ge-3$: $2(x+3)=7+x\Rightarrow x=1$. If $x<-3$: $2(-x-3)=7+x\Rightarrow x=-\tfrac{13}{3}$. Solutions $x=1,\ x=-\tfrac{13}{3}$.

(b) $|2-x|+|x+1|=x+3$, breakpoints $-1,2$. For $x<-1$: $x=-\tfrac23$ (rejected). For $-1\le x<2$: $3=x+3\Rightarrow x=0$. For $x\ge2$: $2x-1=x+3\Rightarrow x=4$. Solutions $x=0,\ x=4$.

(c) $|3+x|\ge -4$: always true. Solution $\forall x\in\mathbb{R}$.

(d) $x|x-1|-3x+3\ge 0$. If $x\ge1$: $(x-1)(x-3)\ge0\Rightarrow x=1\vee x\ge3$. If $x<1$: $x^2+2x-3\le0\Rightarrow -3\le x<1$. Union: $-3\le x\le 1 \ \vee\ x\ge 3$.

$\boxed{\text{(a) } x=1,\,-\tfrac{13}{3};\ \ \text{(b) } x=0,\,4;\ \ \text{(c) } \forall x\in\mathbb{R};\ \ \text{(d) } -3\le x\le1\vee x\ge3}$