Chapter 6.1, Problem 10OE

To determine

To calculate: The simplified form of the expression $\frac{{\left(2y-8\right)}^2}{{\left(2y-8\right)}^3}$ and state restrictions on variable.

Answer to Problem 10OE

The simplified value of the expression $\frac{{\left(2y-8\right)}^2}{{\left(2y-8\right)}^3}$ is $\frac{1}{2y-8}$ and restriction is $y\ne 4$.

Explanation of Solution

Given information:

The expression $\frac{{\left(2y-8\right)}^2}{{\left(2y-8\right)}^3}$.

Formula used:

The simplest form of an algebraic fraction is when numerator and denominator have no common factor other than 1 and $-1$ . Simplification is done by factorizing numerator and denominator.

Calculation:

Consider the expression $\frac{{\left(2y-8\right)}^2}{{\left(2y-8\right)}^3}$.

Recall that the simplest form of an algebraic fraction is when numerator and denominator have no common factor other than 1 and $-1$ . Simplification is done by factorizing numerator and denominator.

So, expand the powers of numerator and denominator and then strike out common terms and simplify it as,

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

To restrict the variable, exclude the value that makes the denominator zero.

So, denominator will be zero when,

  $\begin{array}{c}{\left(2y-8\right)}^3=0\\ 2y-8=0\\ 2y=8\\ y=4\end{array}$

For denominator not to be zero, $y\ne 4$.

Thus, the simplified value of the expression $\frac{{\left(2y-8\right)}^2}{{\left(2y-8\right)}^3}$ is $\frac{1}{2y-8}$ and restriction is $y\ne 4$.