Question

how do I solve this quadratic simultaneous equation 2y+2x=7 and x^2-4y^2=8 it has two sets of solutions and I'm solving by both variables

Answer 1

Sent a picture of the solution to the problem (s).

Answer 2

These are all the possible answers i know from experience.

2x + 2y = 7
x^2 - 4y^2 = 8

2x = -2y + 7
x = -y + 7/2

(-y + 7/2)^2 - 4y^2 = 8

y^2 - 7y + 49/4 - 4y^2 = 8

0 = 3y^2 + 7y + 8 - 49/4

0 = 12y^2 + 28y + 32 - 49

0 = 12y^2 + 28y - 17

y = (-b ± √(b^2 - 4ac))/(2a)

y = (-28 ± √(28^2 - 4(12)(-17)))/(2(12))

Dividing by 2 is like dividing by √(4):

y = (-14 ± √(7*28 + (12)(17)))/(2*6)

y = (-7 ± √(7*7 + (3)(17)))/6

y = (-7 ± 10)/6 = 1/2 or -17/6

x = -y + 7/2 = 3 or 19/3

Solutions are (3,1/2) and (19/3,-17/6)