# How do you combine (4a-2)/(3a+12)-(a-2)/(a+4)?

Jan 8, 2018

See a solution process below:

#### Explanation:

To subtract or add fractions they must be over a common denominator. We can multiply the fraction on the right by the appropriate form of $1$ to put these fractions over a common denominator:

$\frac{4 a - 2}{3 a + 12} - \left(\frac{3}{3} \times \frac{a - 2}{a + 4}\right) \implies$

$\frac{4 a - 2}{3 a + 12} - \frac{3 \left(a - 2\right)}{3 \left(a + 4\right)} \implies$

$\frac{4 a - 2}{3 a + 12} - \frac{3 a - 6}{3 a + 12}$

We can now subtract the numerators over the common denominator:

$\frac{\left(4 a - 2\right) - \left(3 a - 6\right)}{3 a + 12} \implies$

$\frac{4 a - 2 - 3 a + 6}{3 a + 12} \implies$

$\frac{4 a - 3 a - 2 + 6}{3 a + 12} \implies$

$\frac{\left(4 - 3\right) a + \left(- 2 + 6\right)}{3 a + 12} \implies$

$\frac{1 a + 4}{3 a + 12} \implies$

$\frac{a + 4}{3 a + 12}$

We can now factor the numerator and cancel common terms:

$\frac{a + 4}{3 \left(a + 4\right)} \implies$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{a + 4}}}}{3 \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(a + 4\right)}}}} \implies$

$\frac{1}{3}$

Jan 8, 2018

This equals $\frac{1}{3}$, with restriction $a \ne - 4$

#### Explanation:

We can factor the denominator of the left-most expression as $3 \left(a + 4\right)$, thus we multiply the second fraction by $3$.

$= \frac{4 a - 2 - 3 \left(a - 2\right)}{3 a + 12}$

$= \frac{4 a - 2 - 3 a + 6}{3 a + 12}$

$= \frac{a + 4}{3 a + 12}$

$= \frac{a + 4}{3 \left(a + 4\right)}$

$= \frac{1}{3}$

But don't forget that $a \ne - 4$.

Hopefully this helps!