130/x + 180/ (x-5) = 5

make the denominator the same:

[130(x-5)]/[x(x-5)]+[180x]/[x(x-5)]=[130x+180x-650]/[x(x-5)]=[210x-650]/[x^2-5x]

How do I find x once the problem is set up as 310x-650/x^2-5x?

oh my bad, I forgot the other part....

so [210x-650]/[x^2-5x]=5, 210x-650=5x^2-25x, 5x^2-235x+650=0, x^2-47x+130=0, using quadratic formula, we get x=0.5(47-sqrt1689), x=0.5(47+sqrt1689)

hope it helps, thanks!

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