Chapter 6.1, Problem 39WE

To determine

To calculate: The correct simplified form of the expression $\frac{x^2-4y^2}{x-2y}$ when $x=3\text{ and }y=1$ and when $x=4\text{ and }y=2$.

Answer to Problem 39WE

The correct simplified form of the expression $\frac{x^2-4y^2}{x-2y}$ when $x=3\text{ and }y=1$ is $5$ and when $x=4\text{ and }y=2$ the expression cannot be evaluated.

Explanation of Solution

Given information:

The simplification of the expression done by Miguel is,

  $\frac{x^2-4y^2}{x-2y}=x+2y$

When he substituted $x=3\text{ and }y=1$ the result obtained is 5.

When he substituted $x=4\text{ and }y=2$ the result obtained is 8.

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 $\frac{x^2-4y^2}{x-2y}$ done by Miguel as,

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,

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

Substitute $x=3\text{ and }y=1$ in the provided expression.

  $\begin{array}{c}\frac{x^2-4y^2}{x-2xy}=x+2y\\ =3+2\left(1\right)\\ =5\end{array}$

When $x=4\text{ and }y=2$ the expression cannot be evaluated because there is a restriction on the expression. Fractions of the form $\frac{p}{q}$ are defined when $q\ne 0$.

For, $\frac{x^2-4y^2}{x-2y}$ , the denominator should not be equal to 0.

  $\begin{array}{c}x-2y\ne 0\\ x\ne 2y\end{array}$

Since, $4=2\left(2\right)$ , therefore, the expression cannot be evaluated when $x=4\text{ and }y=2$.

Thus, the correct simplified form of the expression $\frac{x^2-4y^2}{x-2y}$ when $x=3\text{ and }y=1$ is $5$ and when $x=4\text{ and }y=2$ the expression cannot be evaluated.