Question

In the following exercises, add or subtract.

$\frac{6y^2+2y-11}{3y-7}+\frac{3y^2-3y+17}{7-3y}$

Short answer

The sum of the expressions is $x-12$.

Step-by-step solution

Step 1. Given Information

The given expression is$\frac{6y^2+2y-11}{3y-7}+\frac{3y^2-3y+17}{7-3y}$

Step 2. Multiply Numerator and denominator of the second fraction by -1 

• Multiply the numerator and denominator of the second term by $-1$.

$\frac{6y^2+2y-11}{3y-7}+\frac{3y^2-3y+17}{7-3y}=\frac{6y^2+2y-11}{3y-7}+\frac{-3y^2+3y-17}{3y-7}$

Step 3. Add the Numerator 

• As the denominator is same for both the terms, add the numerator and place over the common denominator.
• Add the like terms of the numerator.

$\begin{array}{rcl}\frac{6y^2+2y-11}{3y-7}+\frac{-3y^2+3y-17}{3y-7}& =& \frac{6y^2+2y-11-3y^2+3y-17}{3y-7}\\ & =& \frac{3y^2+5y-28}{3y-7}\end{array}$

Step 4. Simplify

• Factor the expression in the numerator and then cancel out the common terms in numerator and denominator.

$\begin{array}{rcl}\frac{3y^2+5y-28}{3y-7}& =& \frac{\cancel{\left(3y-7\right)}\left(y-12\right)}{\cancel{3y-7}}\\ & =& y-12\\ & & \end{array}$

• So, the sum is$y-12$.