Q. 357 - Intermediate Algebra

Question

In the following exercises,translate to a system of equations and solve.

The perimetre of city rectangular park is 1428 feet.The lenght is 78 feet more than twice the width .Find the lenght and width of the park.

Step-by-Step Solution

Answer

The lenght and width of the park are 502 and 212 respectively.

Step 1.Given information

Let l represents length of the park

Let w represents width of the park

Since the perimetre of the rectangular park is 1428 feet.

That means,$2(l+w)=1428$

$$\begin{gathered} (l+w)=\frac{1428}{2} \\ (l+w)=714 \end{gathered}$$

The length is 78 more than twice the width

That means,$l=78+2w$

Step 2.Find the measures of width.

Substitute $l=78+2w$ into the equation $l+w=714$Solve for w.

$$\begin{gathered} l+2w=714 \\ 78+2w+w=714 \\ 78+3w=714 \\ 3w=714-78 \\ 3w=636 \\ w=212 \end{gathered}$$

Therefore,the width of the park area is 212 feets.

Step 3.Find the measures of length.

Substitute $w=212$ into $l=78+2w$ and solve for $l.$

$$\begin{gathered} l=78+2w \\ l=78+2\left(212\right) \\ l=78+424 \\ l=502 \end{gathered}$$

Therefore,the length of the park area is 502 feets