Question

helppp!
solve simultaneous equation:
x=2y-2
x^2=y^2+7?

Answer 1

you just have to substitute the numbers into x. Because it already said x=2y-2, that means:
(2y-2)^2=y^2+7

so you have to do the exponent first:
(2y)²=4y²
(-2)²=4

So now you will have:
4y^2+4=y^2+7

and now you rearrange the unknowns and the numbers:
4y^2-y^2=7-4

and add/subtract:
3y^2=3

now you want the unknown to be by itself so you have to take of the other numbers with the unknown step by step.
So first, take of the 3 by divide it:
y^2=1

And then square root it:
y=√1

and the answer can be -1 or 1

:D

Answer 2

thank you so much!!

Answer 3

no probs ^^