Answer:
Given the system of equations:
[tex]2y-2x =12[/tex] .....[1]
[tex]x^2+y^2=36[/tex] .....[2]
We can write equation [1] as;
2(y-x) = 12
Divide both sides by 2 we get;
[tex]y-x = 6[/tex]
or
y = x+6 ....[3]
Substitute equation [3] in [2] we get;
[tex]x^2+(x+6)^2=36[/tex]
[tex]x^2+x^2+36+12x=36[/tex]
Subtract 36 from both sides we have;
[tex]x^2+x^2+12x=0[/tex]
Combine like terms;
[tex]2x^2+12x =0[/tex]
or
[tex]2x(x+6)=0[/tex]
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)
