x^2 + 2y^2 =17 5x-3y=9 solve x and y

x^2 + 2y^2 =17.....(1)

5x-3y =9.....(2)

We are going to use the substitution method to solve the system

5x-3y = 9

==> x= (3y+9)/5

Now substitute x in (1)

X^2 + 2Y^2 = 17

[(3y+9)/5]^2 +2y^2 = 17

(9y^2 + 54y + 81)/25 + 2y^2 =17

(59/25)y^2 + (54/25)y+ 81/25 =17

(59/25)y^2 + 54/25)y -344/25 =0

Multiply by 25

==> 59y^2 +54y -344=0

==> (y-2)(59y+172)

==> y1= 2   ==> x1= (3y+9)/5 = 15/5 =3

 and y2= -172/59  ==> x2= 0.051 (approx.)