1. ## Inequality

I forgot how to do these things:

$\displaystyle \left|x-2\right|+\left|x+4\right|>6$

Thanks.

2. Originally Posted by Krizalid
I forgot how to do these things:

$\displaystyle \left|x-2\right|+\left|x+4\right|>6$

Thanks.
Consider cases.
If $\displaystyle x\geq 2 \implies |x-2|=x-2\mbox{ and }x<2\implies |x-2|=2-x$

If $\displaystyle x\geq -4 \implies |x+4|=x+4 \mbox{ and }x<-4\implies |x+4|=-x-4$

Thus,
$\displaystyle x<-4\implies (2-x)+(-x-4) >6$
$\displaystyle -4\leq x < 2 \implies (2-x)+(x-4)>6$
$\displaystyle 2\leq x\implies (x+2) + (x-4) > 6$

Now do each one ... carefully.

3. Now I reviewed these things and the procedure is not complicated.

I'll solve this one, may be it may help someone else.

Originally Posted by Krizalid

$\displaystyle \left|x-2\right|+\left|x+4\right|>6$
For $\displaystyle x=-4\implies6>6,$ which is false. For $\displaystyle x=2\implies6>6,$ which is false. Hence, we gotta study the following three cases.
• Case #1: if $\displaystyle x<-4.$ The inequality becomes $\displaystyle (2-x)-(x+4)>6\implies x<-4.$ First solution it's actually $\displaystyle S_1=(-\infty,-4).$
• Case #2: if $\displaystyle -4<x<2.$ The inequality becomes $\displaystyle (2-x)+(x+4)>6\implies6>6,$ which is false, so we have no solution here.
• Case #3: if $\displaystyle x>2.$ The inequality becomes $\displaystyle (x-2)+(x+4)>6\implies x>2.$ Second solution it's actually $\displaystyle S_2=(2,\infty).$

Final solution $\displaystyle S$ will be given by $\displaystyle S=S_{1}\cup S_{2}=(-\infty ,-4)\cup (2,\infty ).$