solve the inequations and provide the solutions in brackets I have 2 inequations which I solve, but when I compare the result with websites like wolframalpha I get different results. 
$|2-x|+|2+x| \leq 10$
I continue with $2-x+2-x\leq10$ OR $2-x+2-x\geq-10$ 
and I get the results $x\geq3$ and $x\leq7 $
$|6-4x|\geq|x-2|$
$6-4x \geq x-2$ OR $6-4x \leq -x+2 $
and I get $x\leq8/5$ and $x\geq4/3$
Are my results correct or not? 
Thank you
 A: You have two critical points (points that evaluate to zero in the absolute values). These are $x=2$ and $x=-2$. Now you have three cases which you have to look at $x\in(-\infty,-2]$, $x\in (-2,2]$ and $x\in (2,\infty)$ [Note: these intervals are important because the expressions in the absolute value functions change sign].
For your second problem the critical points are. $6-4x=0 \to x=\frac{6}{4}$ and $x-2=0 \to x = 2$. What cases do we have to check now?
EDIT: Steps for solving such equations:


*

*write down all terms with absolute values: Here $|2-x|$ and $|2+x|$. 

*find value of x for which these absolute values evaluate to $0$: Here $x=2$ and $x=-2$. These values are called the critical values

*Split the real axis at the critical values ($-2, 2$): Here (a) $x\in(-\infty,-2]$, (b) $x\in (-2,2]$ and (c)$x\in (2,\infty)$

*Look at all cases (a), (b) and (c) separately: Here
(a): First absolute value is positive and the second is negative: $(2-x)-(2+x) \leq 10$. Solving this $-2x \leq 10$ or $x\geq -5$. Compare with (a) $x\in(-\infty,-2]$ to conclude that only values for $x \in [-5,-2]$ are valid solutions for case (a).
(b): First absolute value is positive and the second is positive: $(2-x)+(2+x) \leq 10$. Solving this leads to $4\leq 10$, which is valid for all $x$. Compare with (b) $x\in (-2,2]$ to conclude that only values for  $x\in (-2,2]$ are valid solutions for case (b).
(c): First absolute value is negative and the second is positive: $-(2-x)+(2+x)\leq10$ or $2x\leq 10$ or $x\leq5$. Compare with (c) $x\in (2,\infty)$ to conclude that only $x\in (2,5]$ are valid solutions.
Now combine alle the cases (a), (b) and (c) to conclude that all $x\in [-5,5]$ are solutions to the inequality
I leave the rest to you.
A: You are wrong because you don't manage correctly the sign of the expressions in the absolute values.
In your first inequality you have $ 2-x\ge 0 \iff x\le 2$ and $2+x \ge 0 \iff x\ge -2$. So the inequality  reduces to three systems:
$$
\begin{cases}
x<-2\\
2-x-2-x\le 10
\end{cases}
\lor
\begin{cases}
-2\le x<2\\
2-x+2+x\le 10
\end{cases}
\lor
\begin{cases}
x\ge 2\\
x-2+2+x\le 10
\end{cases}
$$
look well at these systems and understand where they come from, than solve  and find the solution $-5\le x\le 5$ or, in brakets; $[-5,5]$.
Than use the same method for the other inequality.
