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Question
Solve each of the following systems of equations by the method of cross-multiplication
`a^2x + b^2y = c^2`
`b^2x + a^2y = d^2`
Solution
The given system of equations may be written as
`a^2x + b^2y - c^2 = 0`
`b^2x + a^2y - d^2 = 0`
Here,
`a_1 = a^2, b_1 = b^2, c_1 = -c^2`
`a_2 = b^2, b_2 = a^2, c_2 = -d^2`
By cross multiplication, we have
`=> x/(-b^2d^2 + a^2c^2) = (-y)/(-a^2d^2 + b^2c^2) = 1/(a^4 - b^4)`
Now
`x/(-b^2d^2 + a^2c^2) = 1/(a^4 - b^4)`
`=> x = (a^2c^2 - b^2d^2)/(a^4 - b^4)`
And
`(-y)/(-a^2d^2 + b^2c^2) = 1/(a^4 - b^4)`
`=> -y = (-a^2d^2 + b^2c^2)/(a^4 - b^4)`
`=> y = (a^2d^2 - b^2c^2)/(a^4 - b^4)`
Hence `x = (a^2c^2 - b^2d^2)/(a^4 - b^4), y = (a^2d^2 - b^2c^2)/(a^4-b^4)` is the solution of the given system of the equations.
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