What is the algebraic way to solve inequalities like this: $|x+4|<2x$?

Question

What is the algebraic way to solve inequalities like this: $|x+4|<2x$?

Can I rewrite this as $-2x<x+4<2x$ or [$x+4<2x$ and $-(x+4)<2x$]?

I found out that the answer was $x>4$, but how can I do it algebraically instead of by inspection? I also did some algebra that lead me to $x>{-4\over 3}$ when I considered $-(x+4)<2x$, this can't be right.

Answer 1

Your latter rewriting is correct and (in my opinion) the best one. Once you have that

$$\text{both} \; \begin{cases} x+4 < 2x \\ -(x+4) < 2x \end{cases}$$

solve each inequality for $x$. You get

$$\text{both} \; \begin{cases} 4 < x \\ -4/3 < x \end{cases}$$

However, both need to be true, and since $-4/3 < 4$, only the first inequality is necessary to state.

Answer 2

Since $|x+4|\geq 0$ we get $\color{red}{2x> 0}$ so we can square the inequality:

$$ |x+4|^2< 4x^2\implies x^2+8x+16< 4x^2$$

or $$3x^2-8x-16> 0$$

Since $x_1 = 4$ and $x_2 = -{4\over 3}$ so $x\in \color{red}{\mathbb{R}^+\setminus[ -{4\over 3},4]}$.

Answer 3

$$|x+4|<2x \implies 2x\ge 0$$ $$\implies x+4>0 \implies x+4<2x$$ $$\implies x>4$$

conversely

$$x>4\implies 0<x+4<2x $$ $$\implies |x+4|<2x$$

Answer 4

Consider $$f(x)=2x-|x+4| = \begin{cases} 3x+ 4, & x < -4 \\ x-4, & x \ge 4\end{cases}$$ Obviously $f(x)$ is continuous, piecewise linear and strictly increasing, and $f(4)=0$. Therefore $$f(x) > 0 \iff x>4.$$