2x-2a/c^2(p-ax-by)=0,2y-2b/c^2(p-ax-by)=0
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I assume that this is the intended meaning: (2x-2a)/(c^2(p-ax-by))=0, (2y-2b)/(c^2(p-ax-by))=0.

So (2x-2a)/(c^2(p-ax-by))=(2y-2b)/(c^2(p-ax-by)). The denominators cancel out.

Therefore 2x-2a=2y-2b, that is, x-a=y-b, after dividing through by 2.

Also, 2x-2a=0 because, when the numerator is zero the expression is zero and the value of the denominator is irrelevant generally. So 2x=2a, x=a. Similarly, y=b.

Solution: x=a, y=b.

by Top Rated User (1.2m points)

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