Asked by Alicia

Solving Quadratic Systems

16x^2-9y^2=144
y=9-x^2

Please show work I'm having trouble solving it.
Answered by Steve
if you substitute y = 9-x^2, you have

16x^2 - 9(9-x^2)^2 = 144
9x^4 - 178 x^2 + 873 = 0
(x^2-9)(9x^2-97) = 0

Bot those quadratics are easy to solve, no? Each is the difference of two squares.
Answered by Alicia
The correct answer is (3,0) and (3.28,-1.76)


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