Eliminating $x=-3,5$ when we solve the system $y=|3-x|, 4y-(x^2-9)=-24$ $$y=|3-x|$$
$$4y-(x^2-9)=-24$$
I use two methods to solve it,
Method 1: Squaring both sides and eventually get a equation like this$$(x-5)(x-7)(x+3)(x+9)=0$$, $x=-9,-3,5,7$.
Method 2:Using $$y=3-x$$ and $$y=-(3-x)$$, and finally also get $x=-9,-3,5,7$
However, the answers are $x=-9,x=7$ only. 
Is there any method to eliminate $x=-3,x=5$ from the real answers? Or any other methods to avoid adding these 2 wrong solutions into the answers?


Is there any method to eliminate $x=-3,x=5$ from the real answers?

For method 1 : 
You have to have
$$x^2-33=4y=4|3-y|\ge 0\implies x^2\ge 33$$
For method 2 : 
As f'' comments, note that $|3-x|=3-x$ holds only when $3-x\ge 0,$ i.e. $x\le 3$ and that $|3-x|=-(3-x)$ holds only when $3-x\lt 0$, i.e. $x\gt 3$.
So, if you have $y=3-x$, then you get
$$4(3-x)-(x^2-9)=-24\implies x=5,-9$$
with $x\le 3$. Therefore, $x=-9$ is the only solution in this case.
