Question

In the following exercises, perform indicated operations.

$\frac{2y}{y+3}+\frac{3}{y-1}$

Short answer

The resultant expression is$\frac{2y^2+y+9}{(y+3)(y-1)}$.

Step-by-step solution

Step 1. Given Information

The given expression is$\frac{2y}{y+3}+\frac{3}{y-1}$.

Step 2. Express with common denominator   

• The LCD of the terms of given expression is $(y+3)(y-1)$.
• Multiply the numerator and denominator of first term by $(y-1)$and second term by $(y+3)$.

role="math" localid="1645168052631" $\frac{2y(y-1)}{(y+3)(y-1)}+\frac{3(y+3)}{(y-1)(y+3)}=\frac{2y^2-2y}{(y+3)(y-1)}+\frac{3y+9}{(y-1)(y+3)}$

Step 3. Simplify

• As the denominator is the same for both the terms, perform addition in the numerator and place it over the common denominator.

$\frac{2y^2-2y+3y+9}{(y+3)(y-1)}=\frac{2y^2+y+9}{(y+3)(y-1)}$

• So, the resultant expression is $\frac{2y^2+y+9}{(y+3)(y-1)}$.