Question

Which is one solution of the following system
2y-2x=12
x^2+y^2+36

Answer 1

A.) (-6,0)

Explanation:

It's on edge.

Answer 2

Given the system of equations:

`2y-2x =12` .....[1]

`x^2+y^2=36` .....[2]

We can write equation [1] as;

2(y-x) = 12

Divide both sides by 2 we get;

`y-x = 6`

or

y = x+6 ....[3]

Substitute equation [3] in [2] we get;

`x^2+(x+6)^2=36`

`x^2+x^2+36+12x=36`

Subtract 36 from both sides we have;

`x^2+x^2+12x=0`

Combine like terms;

`2x^2+12x =0`

or

`2x(x+6)=0`

By zero product property:

x =0 and x+6 = 0

or

x = 0 and x = -6

now, substitute the given values of x in [3] we have;

for x = 0

y = 0+6 = 6

for x = -6

y = -6 + 6 = 0

Therefore, the solution for the given system of equation is, either (0, 6) or (-6, 0)