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multiply each term by x(x+5), this gets rid of the denominators and we have
(x+5) + x = x(x+6)
x + 5 + x = x^2 + 6x
2x + 5 = x^2 + 6x
Arrange as a quadratic equation on the right
0 = x^2 + 6x - 2x - 5
x^2 + 4x - 5 = 0
Factors to
(x+5)(x-1) = 0
Two solutions
x = -5
x = 1
When you try these solutions in the original problem, you can see that only
x = 1 is a valid solution, x=-5 gives us denominators of 0