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Find two numbers whose ratio is 4 to 3 and are such that the sum of their squares is 100.
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8 and 6 which squares at 64 and 36 to have a sum of 100.& I can't think of any others. .
(x/y)=(4/3) x^2+y^2=100 x=4y/3 (4y/3)^2+y^2=100 (16y^2)/9+y^2=100 y^2(16/9+1)=100 y^2=36 y=6 or y=-6 Plugging back into the equation x/y=4/3 x/6=4/3 x=8 or x/-6=4/3 x=-8 So there are two sets of numbers that satisfy the parameters. 6 and 8, or -6 and -8.
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