Math
posted by Alexander on .
Find the solution set of the equation
{x^2+y^2=34
{x+y=2

just substitute y=2x and you have
x^2 + (2x)^2 = 34
2x^2  4x  30 = 0
x^2  2x  15 = 0
now it should be clear ... 
y = x +2
x^2 + (x +2)^2 = 34
x ^2 + x^2 2x 2x +4 = 34
2x^2 4x + 4 = 34
2x^ 4x + 4 34 = 3434
2x^2 4x  30 = 0
2(x^2 2x 15) =0
2(x5)(x+3) =0
x 5 =0
x =5
x+ 3 = 0
x =3
y = 3 ,5
(5,3), (3,5)