# How do you solve x+(x+5)/(x-7)=(10x-58)/(x-7) and check for extraneous solutions?

May 20, 2017

#### Explanation:

First of all, seeing how this question deals with fractions, multiply the value of $x$ by $\frac{x - 7}{x - 7}$ to get $\frac{{x}^{2} - 7 x}{x - 7}$.

Now we have three fractions with the same denominator:

$\frac{{x}^{2} - 7 x}{x - 7} + \frac{x + 5}{x - 7} = \frac{10 x - 58}{x - 7}$

Because the denominators are all the same, they eventually cancel out by cross-multiplication, so we can drop them to get a simple equation:

$\left({x}^{2} - 7 x\right) + \left(x + 5\right) = 10 x - 58$

At this point, it is a matter of combining like terms and simplifying for an answer

$\left({x}^{2} - 7 x\right) + \left(x + 5\right) = 10 x - 58$
=> ${x}^{2} - 6 x + 5 = 10 x - 58$
=>${x}^{2} - 16 x + 63 = 0$
=> $\left(x - 9\right) \left(x - 7\right) = 0$
$x = 7 , 9$

Now we have two answers, and the problem seems over, right? Well, it isn't, as we have an extraneous solution: 7. This is because, when plugged back into the original equation, 7 creates a denominator of 0; dividing by zero is impossible, so we exclude 7 as a possible answer. So we are left with one final answer: 9