# How do you combine 4/(3p)-5/(2p^2)?

Oct 17, 2017

$\frac{8 p - 15}{6 {p}^{2}}$

#### Explanation:

First get common denominators.

$3 p$ and $2 {p}^{2}$ will have the least common multiple of $6 {p}^{2}$

We can find this by looking at one term, determining what factors it is missing from the other term, and then multiplying those factors in.

$3 p = 3 \cdot p$
$2 {p}^{2} = 2 \cdot p \cdot p$

They both have a $p$, so let's take the first one and give it the ones the second one has except for a $p$.
$3 \cdot p \textcolor{b l u e}{\cdot 2 \cdot p} = 6 {p}^{2}$

Now we multiply each fraction by a unit factor to get the common denominator.

$\frac{4}{3 p} - \frac{5}{2 {p}^{2}}$

$\textcolor{b l u e}{\frac{2 p}{2 p}} \cdot \frac{4}{3 p} - \textcolor{b l u e}{\frac{3}{3}} \cdot \frac{5}{2 {p}^{2}}$

$\frac{8 p}{6 {p}^{2}} - \frac{15}{6 {p}^{2}}$

Now we can finally combine the fractions

$\frac{8 p - 15}{6 {p}^{2}}$

Oct 17, 2017

$\frac{8 p - 15}{6 {p}^{2}}$

#### Explanation:

Take L CM for the denominator:

Factors of 3p are $3 , \textcolor{red}{p}$
Factors of $2 {p}^{2}$ are $2 , p , \textcolor{red}{p}$

L C M of the Denominator is $3 \cdot 2 \cdot p \cdot \textcolor{red}{p} = 6 {p}^{2}$
$\textcolor{red}{p}$ is used only once as it is appearing in both.

Combining the two terms,
$\left(\frac{4}{3 p}\right) - \left(\frac{5}{2 {p}^{2}}\right) = \frac{\left(4 \cdot 2 \cdot p\right) - \left(5 \cdot 3\right)}{6 {p}^{2}}$
$= \frac{8 p - 15}{6 {p}^{2}}$