Q7. - Algebra 1

Question

Corey has 42 feet of fencing around his garden. The garden is rectangular in shape, and its length is equal to twice the width minus 3 feet. Define the variables, and write a system of equations to find the length and width of the garden. Solve the system by using substitution.

Step-by-Step Solution

Answer

Let $l$ be the length of the rectangular garden and $\mathit{w}$ be the width of the rectangular garden.

The system of equations to find the length and width of the garden is:

$\begin{array}{c}l+w=21\text{ ..... }\left(1\right)\\ l=2w-3\text{ .......}\left(2\right)\end{array}$ 

The length and width of the garden are 13 feet and 8 feet respectively.

Step 1. Define the variables and write a system of equations to find the length and width of the garden.

Let $l$ be the length of the rectangular garden and be the width of the rectangular garden.

It is given that garden has 42 feet of fencing. That implies the perimeter of the garden is 42 feet.

It is known that:

$P=2(l+w)$

Where $P$ is perimeter, $l$ is the length and is the width of the rectangle.

Therefore,

$\begin{array}{l}P=2(l+w)\\ 42=2(l+w)\\ 21=l+w\text{ .....}\left(1\right)\end{array}$

It is given that length is equal to twice the width minus 3 feet.

Therefore,

$l=2w-3.....\left(2\right)$

The system of equations to find the length and width of the garden is:

$\begin{array}{c}l+w=21\text{ ....}\left(1\right)\\ l=2w-3\text{ ....}\left(2\right)\end{array}$

Step 2. Solve the given system of equations.

Substitute the value of $l$ from the equation (2) in the equation (1).

$\begin{array}{c}l+w=21\\ 2w-3+w=21\\ 3w=21+3\\ 3w=24\\ w=\frac{24}{3}\\ w=8\end{array}$

The value of $w$ is 8.

Substitute 8 for $w$ in the equation (2) to find the value of $l$.

$\begin{array}{c}l=2w-3\\ =2\left(8\right)-3\\ =16-3\\ =13\end{array}$

The value of $l$ is 13.

Therefore, the length and width of the garden are 13 feet and 8 feet respectively