Chapter A.3, Problem 57E

Solution

Simplify.

  $\frac{x}{4(x-3)}$ .

Given:

The given fraction is $\frac{\frac{2x^{2}y}{{(x-3)}^2}}{\frac{8xy}{x-3}}$ .

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 $\frac{\frac{2x^{2}y}{{(x-3)}^2}}{\frac{8xy}{x-3}}$ .

  $\begin{array}{c}\frac{\frac{2x^{2}y}{{(x-3)}^2}}{\frac{8xy}{x-3}}=\frac{2x^{2}y(x-3)}{{(x-3)}^{2}8xy}\\ =\frac{2x^{2}y(x-3)}{(x-3)(x-3)8xy}\\ =\frac{2x^{2}y}{(x-3)8xy}\\ =\frac{x}{(x-3)4}\\ =\frac{x}{4(x-3)}\end{array}$

Thus by simplifying $\frac{\frac{2x^{2}y}{{(x-3)}^2}}{\frac{8xy}{x-3}}$ to a single fraction it comes $\frac{x}{4(x-3)}$ .