5 y + 7 x = 31 x y
5 y + 7 x = 29 x y
well, (0,0) would work but then we would be dividing by zero in the original.
5/x+7/y=31,7x+5y=29xy by comparision and subsitition method
2 answers
using substitution,
7/y = 31-5/x so
y = 7/(31-5/x) = 7x/(31x-5)
7x+5y=29xy
y = -7x/(5-29x)
7x + 35x/(31x-5) = 29x * 7x/(31x-5)
7x(31x-5) + 35x = 203x^2
14x^2 = 0
As Damon noted, x=0 is not allowed, so there are no solutions.
If you plot the difference between the two functions, you can see that it is zero only at x=0.
http://www.wolframalpha.com/input/?i=7x%2F%2831x-5%29%2B7x%2F%285-29x%29+for+x%3D+-0.1+to+0.1
7/y = 31-5/x so
y = 7/(31-5/x) = 7x/(31x-5)
7x+5y=29xy
y = -7x/(5-29x)
7x + 35x/(31x-5) = 29x * 7x/(31x-5)
7x(31x-5) + 35x = 203x^2
14x^2 = 0
As Damon noted, x=0 is not allowed, so there are no solutions.
If you plot the difference between the two functions, you can see that it is zero only at x=0.
http://www.wolframalpha.com/input/?i=7x%2F%2831x-5%29%2B7x%2F%285-29x%29+for+x%3D+-0.1+to+0.1
2 answers