# How do you subtract 3/(7b)-8/(3b^2)?

Oct 28, 2016

$\frac{9 b - 56}{21 {b}^{2}}$

#### Explanation:

$\frac{3}{7 b} - \frac{8}{3 {b}^{2}}$

First find the common denominator.

The least common multiple of $3 \mathmr{and} 7$ is $21$.

The least common multiple of $b$ and ${b}^{2}$ is ${b}^{2}$.

So, the common denominator is $21 {b}^{2}$.

Multiply the first term by $\frac{3 b}{3 b}$ and the second term by $\frac{7}{7}$.

$\frac{3 b}{3 b} \cdot \frac{3}{7 b} - \frac{7}{7} \cdot \frac{8}{3 {b}^{2}}$

$\frac{9 b}{21 {b}^{2}} - \frac{56}{21 {b}^{2}}$

$\frac{9 b - 56}{21 {b}^{2}}$

Oct 28, 2016

$\frac{3}{7 b} - \frac{8}{3 {b}^{2}}$

$= \frac{3}{7 b} \cdot 1 - \frac{8}{3 {b}^{2}} \cdot 1$

$= \frac{3}{7 b} \cdot \frac{3 b}{3 b} - \frac{8}{3 {b}^{2}} \cdot \frac{7}{7}$

$= \frac{9 b}{21 {b}^{2}} - \frac{56}{21 {b}^{2}}$

$= \frac{9 b - 56}{21 {b}^{2}}$

The fraction above cannot be simplified, so this is the fraction you'd be looking to get if you were to subtract the second fraction (right) from the first fraction (left).