2x^2-2y=y^2=4x-y=8?

Question

i know the answer is x=2,y=0, please PLEASE i need to know the working

Answer 1

There is something wrong with the initial expressions.
If x=2, y = 0, then y^2=0 and not 8 as the equalities seems to indicate.
Try to verify the equations. May be there is a system of two equations.
For instance, if the equations were
2x^2-2y=8
4 x -y =8
then x =2, y = 0 solves the system. The working will be
y = 4 x - 8 from the second, and substituting y in the first
2 x^2 - 8 x + 16 = 8, so we have a second order equation
x^2 - 4 x +4=0 whose roots are x = (4 +- 0)/2 = 2, and then
y = 4 x 2 - 8 =0.
I apologize for my bad english, I am not english speaker.

Answer 2

It's a simple simultaneous equation.
2x^2-2y=y^2
4x-y=8

I would get y on it's own.
so 4x-8=y
Sub that in.
2x^2-2(4x-8)=(4x-8)(4x-8)
2x^2-8x+16 = 16x^2 - 64x + 64
(move them to the right)
14x^2 - 56x + 48 = 0
make it easier
7x^2 - 28x + 24 = 0
now to factorise...ok so it doesn't factorise which is a bother.
well you gotta do complete the square or use the quadratic formula. Complete the square gave me 2+-root 4/7 which is close to 2...

Answer 3

4x-y=8 mult.by (-2)

-8x+2y= -16
add 2x^2-2y= 8
-------------------------
2x^2-8x=-8 2x^2-8x+8=0


x^2-4x+4=0 (x- 2)^2=0
(x-2)=0 so x=2 sub in any equation for example 4x-y=8
so 8-y=8 so y=0