# Simplify. `1/(y^2-y-2) - 1/(y^2+2)`1 - 1 ______ ________ y^2-y-2. y^2+2

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### 1 Answer

`1/(y^2-y-2) - 1/(y^2+2)`

Since the two fraction have unlike denominators, we need to determine its LCD.

The first denominator has a factor of `(y-2)(y+1)` while the second denominator is not factorable. So the LCD is the product of the two denominators which is `(y^2-y-2)(y^2+2)` .

Then, multiply the first fraction by `(y^2+2)/(y^2+2)` and the second fraction by `(y^2-y-2)/(y^2-y-2)` in order for the fractions to have a same denominator.

=`1/(y^2-y-2)* (y^2+2)/(y^2+2) - 1/(y^2+2)*(y^2-y-2)/(y^2-y-2)`

`=(y^2+2)/((y^2-y-2)(y^2+2)) - (y^2-y-2)/((y^2-y-2)(y^2+2))`

Now that the fractions have same denominators, proceed to subtract.

`= (y^2+2 -(y^2-y-2))/((y^2-y-2)(y^2+2))`

`=(y^2+2-y^2+y+2)/((y^2-y-2)(y^2+2))`

`=(y+4)/((y^2-y-2)(y^2+2))`

**Hence, `1/(y^2-y-2)-1/(y^2+2)=(y+4)/((y^2-y-2)(y^2+2))` .**