Solving an absolute-value inequality: $−|x|+2 \geq 8x $

Question

How would I go about solving the domain of this inequality? $$−|x|+2 \geq 8x $$ I can't combine the $x$'s so I don't know what to do.

Could I say: $$-x + 2 ≥ 8x $$ and $$x - 2 ≥ 8x $$ and solve the domain from there?

Answer 1

case 1: if $x\ge0$

$$-x+2\ge 8x$$ $$2\ge 9x$$ we get $$\frac{2}{9}\ge x$$ but we already assumed $x\ge0$ thus $$\frac{2}{9}\ge x\ge 0$$ case 2: if $x<0$ $$x+2\ge 8x$$ $$2\ge 7x$$

$$\frac{2}{7}\ge x$$

but we already assumed $x<0$ thus we get $$x<0$$

Take the union of both solutions

Answer 2

We can write $$2\geq 8x+|x|$$. If we have $$x\geq 0$$ we have $$2\geq 8x+x$$ If $$x<0$$ then we get $$2\geq 8x-x$$ Can you finish?

Answer 3

The case analysis gives

$$\begin{cases}x\ge0\to-x+2\ge8x,\\x\le0\to x+2\ge8x\end{cases}$$

or

$$\begin{cases}x\ge0\to x\le\frac29,\\x\le0\to x\le\frac27.\end{cases}$$

Combining the inequations, finally

$$x\le\frac29.$$