Chapter A.3, Problem 55E

Solution

Simplify.

  $\frac{3\left(x-3\right)}{28}$ .

Given:

The given fraction is $\frac{x^2-3x}{14y}÷\frac{2xy}{3y^2}$ .

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 $\frac{x^2-3x}{14y}÷\frac{2xy}{3y^2}$ .

  $\begin{array}{c}\frac{x^2-3x}{14y}÷\frac{2xy}{3y^2}=\frac{x^2-3x}{14y}\times \frac{3y^2}{2xy}\\ =\frac{x(x-3)(3y^2)}{28x{y}^2}\\ =\frac{(x-3)(3)}{28}\\ =\frac{3\left(x-3\right)}{28}\end{array}$

Thus by simplifying $\frac{x^2-3x}{14y}÷\frac{2xy}{3y^2}$ to a single fraction it comes $\frac{3\left(x-3\right)}{28}$ .