Question 271534
Let the fractions be x & y
xy=3/32
x+y=7/8
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y=(7/8)-x
x((7/8)-x)=3/32 substitute (7/8)-x for y.
32x((7/8)-x)=3
28x-32x^2=3
32x^2-28x+3=0
x=3/4, y=1/8 or vice versa
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Ed
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*[invoke quadratic "x", 32, -28, 3]