a^2x+b^2y=c^2

b^2x+a^2y=d^2
in Algebra 1 Answers by Level 1 User (140 points)

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Multiply the first equation by b^2 and the second by a^2 and we get: a^2b^2x+b^4y=b^2c^2 and a^2b^2x+a^4y=a^2d^2. We can eliminate the x term by subtracting one equation from the other: y(b^4-a^4)=b^2c^2-a^2d^2, from which y=(b^2c^2-a^2d^2)/(b^4-a^4) or (a^2d^2-b^2c^2)/(a^4-b^4). We could substitute this value of y into one of the original equations, but because of the symmetry of the equations we could simply repeat this process by multiplying the first equation by a^2 and the second by b^2 to eliminate the y term through subtraction. When we do this we get: x(a^4-b^4)=a^2c^2-b^2d^2, making x=(a^2c^2-b^2d^2)/(a^4-b^4) or (b^2d^2-a^2c^2)/(b^4-a^4).

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