# How do you simplify (3+y)/(y^2-9) -( y-3)/(9-y^2)?

Jul 23, 2016

$= \frac{2 y}{{y}^{2} - 9}$

#### Explanation:

In adding and subtracting fractions, we need to have a common denominator.

The denominators are almost the same, but the signs in the second fraction are the wrong way around. We therefore need to do a 'switch-round'.

Note: $\left(2 x - 3 y + 4 z\right) = - \left(- 2 x + 3 y - 4 z\right)$ If the sign in the front of a bracket changes, the signs inside the bracket change,

$\frac{3 + y}{{y}^{2} - 9} - \frac{y - 3}{9 - {y}^{2}} \text{ can be written as } \frac{3 + y}{{y}^{2} - 9} + \frac{y - 3}{{y}^{2} - 9}$

Now we have the same denominator and can add the fractions.

$\frac{3 + y + y - 3}{{y}^{2} - 9} = \frac{2 y}{{y}^{2} - 9}$

An alternative method would be to factorise first:

$\frac{\left(3 + y\right)}{\left(y + 3\right) \left(y - 3\right)} - \frac{\left(y - 3\right)}{\left(3 + y\right) \left(3 - y\right)}$

We still need to do a switch round in the second fraction.
(The signs change in only ONE bracket)

$\frac{{\cancel{3 + y}}^{1}}{\cancel{y + 3} \left(y - 3\right)} + \frac{{\cancel{y - 3}}^{1}}{\left(3 + y\right) \cancel{y - 3}}$

Converting to a common denominator .

$\frac{1}{\left(y - 3\right)} + \frac{1}{\left(3 + y\right)} = \frac{3 + y + y - 3}{\left(3 + y\right) \left(y - 3\right)}$

$= \frac{2 y}{{y}^{2} - 9}$