Question

How do you solve the system `x^2-2x+2y+2=0` and `-x^2+2x-y+3=0`?

Answer 1

`(-2,-5),(4,5)`

Explanation:

`x^2-2x+2y+2=0to(1)`

`-x^2+2x-y+3=0to(2)`

`"add equations "(1)" and "(2)" term by term"`

`(x^2-x^2)+(-2x+2x)+(2y-y)+5=0`

`y+5=0rArry=-5`

`"substitute "y=-5" into either "(1)" or "(2)" and"`
`"solve for x"`

`(1)tox^2-2x-10+2=0`

`x^2-2x-8=0larrcolor(blue)"in standard form"`

`(x-4)(x+2)=0`

`x-4=0rArrx=4`

`x+2=0rArrx=-2`

`"solutions are "(-2,-5)" or "(4,-5)`
graph{(x^2-2x+2y+2)(x^2-2x+y-3)=0 [-10, 10, -5, 5]}