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
2 answers:
Sent a picture of the solution to the problem (s).
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
<span>Solutions are (3,1/2) and (19/3,-17/6)</span>
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<em>The answer is 5s, Hope this helps for your work. </em>
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since f(-x)=f(x)
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