Chapter A.3, Problem 51E

Solution

Simplify.

  $\frac{1}{y}$ .

Given:

The given fraction is $\frac{2y^2+9y-5}{y^2-25}\cdot \frac{y-5}{2y^2-y}$ .

Concept Used:

• Remove any grouping symbol such as brackets and parentheses by multiplying factors.
• Use the exponent rule to remove grouping if the terms are containing exponents.
• Combine the like terms by addition or subtraction.
• Combine the constants.

Calculation:

The given fraction is $\frac{2y^2+9y-5}{y^2-25}\cdot \frac{y-5}{2y^2-y}$ .

  $\begin{array}{c}\frac{2y^2+9y-5}{y^2-25}\cdot \frac{y-5}{2y^2-y}=\frac{2y^2+9y-5}{y^2-5^2}\cdot \frac{y-5}{y(2y-1)}\\ =\frac{2y^2+9y-5}{(y+5)(y-5)}\cdot \frac{y-5}{y(2y-1)}\\ solve2y^2+9y-5=2y^2+(10-1)y-5\\ =2y^2+10y-y-5\\ =2y(y+5)-1(y+5)\\ =(2y-1)(y+5)\\ putthevalueabove\\ \frac{2y^2+9y-5}{y^2-5^2}\cdot \frac{y-5}{y(2y-1)}=\frac{(2y-1)(y+5)}{(y+5)(y-5)}\cdot \frac{y-5}{y(2y-1)}\\ =\frac{(2y-1)(y+5)(y-5)}{y(2y-1)(y+5)(y-5)}\\ =\frac{1}{y}\end{array}$

Thus by simplifying $\frac{2y^2+9y-5}{y^2-25}\cdot \frac{y-5}{2y^2-y}$ to a single fraction it comes $\frac{1}{y}$ .