I need to solve this system of equations using algebra. I've tried splitting the equations but I always got results like 0=0 and values for x which contradicted each other. I know the solution is [-2, 2] but I can't find the way to solve it.

{|x-2| = 2-x

{|x+2| = x+2

I'm sorry for the formatting, don't think it's possible to type accolades for multiple lines.

  1. $|x-2|=2-x \iff 2-x \ge 0 \iff x \le 2$:

  2. $|x+2|=x+2 \iff x+2 \ge 0 \iff -2 \le x$.


$|x-2|=2-x $ and $|x+2|=x+2 \iff x \in [-2,2].$


Whenever an expression of the type $|f(x)|$ appears, you should separate two cases: $f(x)\geq0$ and $f(x)\leq0$.

In your case, because you have an expression $|x-2|$, you should separate the cases when $x-2\geq0$ (i.e. when $x\geq2$) and when $x-2\leq0$ (i.e. when $x\leq2$).

Case 1:

$x\geq2$. In this case, $x-2\geq0$ meaning the equation $|x-2|=2-x$ becomes $x-2=2-x$ which simplifies to $2x=4$ and $x=2$. So, the first equation in case $1$ holds if and only if $x=2$.

The second equation is even simpler. Because $x\geq 2$, it is clear that $x+2\geq 4>0$, so $|x+2|=x+2$ and the second equation becomes $x+2=x+2$ which is always true.

So, in Case 1, $x=2$ is the only solution that satisfies both equations.

Case 2:

$x\leq 2$ means $|x-2|=2-x$ is always true because $|x-2|=-(x-2)=2-x$. So, the first equation is always true.

The second equation is unclear, and you have to split two additional cases:

Case 2.1:

$x\geq -2$. Follow the steps from Case 1

case 2.2:

$x\leq -2$. Follow the steps from Case 1.


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