Solve the system

[ozero]

{ | x - 2 | + y^2 = 2-x
{ y=x^2+2x-15

(I don't know what I can do with first equation, if second turn to y=(x+5)(x-3))

 

[Deleted member 4993]

{ | x - 2 | + y^2 = 2-x
{ y=x^2+2x-15

(I don't know what I can do with first equation, if second turn to y=(x+5)(x-3))

{ | x - 2 | + y^2 = 2-x
{ y=x^2+2x-15
............................... (1)

{ y=x^2+2x-15.............................................(2)

If x < 2

Eqn. (1) becomes

{ | x - 2 | + y^2 = 2 - x

- (x - 2) + y^2 = 2 - x → y^2 = 0 ........ continue.

 

[skeeter]

[math]|x-2| = (x-2) \text{ if } x \ge 2[/math]
[math]|x-2| = -(x-2) = (2-x) \text{ if } x < 2[/math]
see what you can do from here ...

 

[lex]

You end up with two questions to do:

[MATH]\boxed{x-2 + y^2 = 2-x \hspace{2cm} \text{when }x\gt2\\ y = (x + 5)(x - 3)}[/MATH]and

[MATH]\boxed{2-x + y^2 = 2-x \hspace{2cm} \text{when }x\le 2\\ y = (x + 5)(x - 3)}[/MATH]
i.e. solve

(1)
[MATH]\boxed{y^2 = 2(2-x) \hspace{2cm} \text{when }x\gt2\\ y = (x + 5)(x - 3)}[/MATH]and

(2)
[MATH]\boxed{y^2 = 0\hspace{2cm} \text{when }x\le 2\\ y = (x + 5)(x - 3)}[/MATH]
Note for (1), the implication of [MATH]y^2=2(2-x) \text{ when } x\gt 2[/MATH]