$\displaystyle \frac {x}{x+1} + \frac {5}{x}={1}{x^2+x}$
Let's combine the two fractions on the left by using the common denominator of (x+1)(x) [simply the product of the two denominators].
This gives:
$\displaystyle \frac{x^2}{x^2 + x} + \frac{5x + 5}{x^2 + x} = \frac{1}{x^2+x}$
$\displaystyle = \frac{x^2 + 5x + 5}{x^2 + x} = \frac{1}{x^2+x}$.
Since the denominators are both the same, the only way the fractions can be equal is if the numerators are ALSO the same. So, now we don't need to worry about the denominators. We just need to solve:
$\displaystyle x^2 + 5x + 5 = 1$
$\displaystyle x^2 + 5x + 4 = 0$
$\displaystyle (x + 4) (x + 1) = 0$
so x = -4 or x = -1.
Lastly, we need to make sure that, if we plug these values in for x, we don't get zero in the denominator of our original problem.
Note that if we plug in x = -1 in the original problem, in the first fraction, we get zero in the denominator. Thus, x = -1 is NOT a valid solution, as it isn't in the domain.
x = -4 doesn't cause any problems, so that's our only answer.
1) Get a common denominator.
2) Combine so you have 1 fraction on each side.
3) Cross-multiply (or, like in this case, if either both numerators or both denominators are the same, you can just throw them out).
4) Solve the remaining polynomial using factoring, quadratic formula, whatever method works best.