# How do you solve  3/x = 12/(x+7)?

Sep 27, 2015

The solutions is $x = \frac{7}{3}$.

#### Explanation:

An identity holds if and only if the identity between the inverses hold (as long as you don't have something like $0 = 0$ of course). So, in you case, you have that

$\frac{3}{x} = \frac{12}{x + 7} \setminus \iff \frac{x}{3} = \frac{x + 7}{12}$

The only thing we have to take care about is to make sure that the original denominators aren't zero, i.e. $x \setminus \ne 0$ and $x + 7 \setminus \ne 0$. This means that, when we'll eventually find solutions, we have to make sure that these solutions are nor $0$ nor $- 7$.

Now let's solve the equality between inverses:

$\frac{x}{3} = \frac{x + 7}{12} \setminus \iff 12 x = 3 \left(x + 7\right)$.

Expanding, we have $12 x = 3 x + 21$. Isolating the $x$-terms and the costants, we get $9 x = 21$, and thus we can solve for $x$: $x = \frac{21}{9}$, which you can simplify (dividing by $3$ both numerator and denominator) into $\frac{7}{3}$.