Chapter 6.1, Problem 40WE

To determine

To calculate: The correct simplified form of the expression $\frac{x^2-4y^2}{x^2-2xy}$.

Answer to Problem 40WE

The correct simplified form of the expression $\frac{x^2-4y^2}{x^2-2xy}$ is $\frac{x+2y}{x}$.

Explanation of Solution

Given information:

The simplification of the expression done by Donna is,

  $\begin{array}{c}\frac{x^2-4y^2}{x^2-2xy}=\frac{{\cancel{x}}^{\cancel{2}}-4y^2}{{\cancel{x}}^{\cancel{2}}-2xy}\\ =\frac{-4y^2}{-2xy}\\ =\frac{2y}{x}\end{array}$

Formula used:

The difference of square of two numbers a and bis $a^2-b^2=\left(a-b\right)\left(a+b\right)$.

Calculation:

Consider the simplification of the expression done by Donna as,

  $\begin{array}{c}\frac{x^2-4y^2}{x^2-2xy}=\frac{{\cancel{x}}^{\cancel{2}}-4y^2}{{\cancel{x}}^{\cancel{2}}-2xy}\\ =\frac{-4y^2}{-2xy}\\ =\frac{2y}{x}\end{array}$

Terms cannot be cancelled like this. Because $x^2-4y^2$ is one entire term and $x^2-2xy$ is entire second term. The term $x^2-4y^2$ is divided by $x^2-2xy$.

Recall that difference of square of two numbers a and bis $a^2-b^2=\left(a-b\right)\left(a+b\right)$.

Apply it,

  $\begin{array}{c}\frac{x^2-4y^2}{x^2-2xy}=\frac{x^2-{\left(2y\right)}^2}{x\left(x-2y\right)}\\ =\frac{\left(x+2y\right)\left(x-2y\right)}{x\left(x-2y\right)}\end{array}$

Strike off the similar terms,

  $\begin{array}{c}\frac{x^2-4y^2}{x^2-2xy}=\frac{x^2-{\left(2y\right)}^2}{x\left(x-2y\right)}\\ =\frac{\left(x+2y\right)\left(x-2y\right)}{x\left(x-2y\right)}\\ =\frac{x+2y}{x}\end{array}$

Substitute $x=4\text{ and }y=1$ in the provided expression.

  $\begin{array}{c}\frac{x^2-4y^2}{x^2-2xy}=\frac{4^2-4{\left(1\right)}^2}{4^2-2\left(4\right)1}\\ =\frac{16-4}{16-8}\\ =\frac{12}{8}\\ =\frac{3}{2}\end{array}$

Substitute $x=4\text{ and }y=1$ in the expression solved by Donna,

  $\begin{array}{c}\frac{x^2-4y^2}{x^2-2xy}=\frac{2y}{x}\\ =\frac{2\left(1\right)}{4}\\ =\frac{1}{2}\end{array}$

Therefore, Donna’s simplification is wrong.

Thus, the correct simplified form of the expression $\frac{x^2-4y^2}{x^2-2xy}$ is $\frac{x+2y}{x}$.