Q. 4.40 - Intermediate Algebra

Question

Translate to a system of equations and then solve:

Alexis wants to build a rectangular dog run in her yard adjacent to her neighbor’s fence. She will use $136$feet of fencing to completely enclose the rectangular dog run. The length of the dog run along the neighbor’s fence will be $16$ feet less than twice the width. Find the length and width of the dog run.

Step-by-Step Solution

Answer

The length and width of the rectangular dog run is $40,28$ feet respectively.

Step 1. Given

We know that total fencing used is $136$ feet. 
The length of the dog run is$16$ feet less than twice the width.

Step 2. Explanation

Assuming $L,W$ to be the length and width of the dog run respectively, total fencing used is $136$ feet, i.e.,$2L+2W=136$.
This can also be written as $L+W=68$ by taking $2$ common from the equation.
And as the length is$16$ feet less than the width, it can be written as$L=2W-16$.
Substituting this value of length in the first equation will give us the length and width of the rectangular dog run.

Step 3. Finding the width

We have $L=2W-16$.
We will substitute this in $L+W=68$.

It will give us $(2W-16)+W=68$.

Solving the equation out gives us,
$$\begin{gathered} 3W-16=68 \\ 3W=84 \\ W=28 \end{gathered}$$

This shows that width of the rectangular dog run is $28$ feet.

Step 4. Finding the length

As we now know that width is $28$ feet, we can calculate the length by substituting value of width in $L+W=68$.

Substituting the value will give us,
$$\begin{gathered} L+28=68 \\ L=40 \end{gathered}$$

This shows that length of the rectangular dog run is $40$ feet.

Step 5. Check the solution

We got the length and width of the dog run as $40,28$ feet. We also know that $2L+2W=136$. We will put the values in the equation and if the equation satisfies or becomes true, our answer is right.

So,

$$\begin{gathered} 2L+2W=136 \\ 2\left(40\right)+2\left(28\right)=136 \\ 80+56=136 \\ 136=136 \end{gathered}$$

Hence the equation is true.