# If the length L of a rectangle is 3 meters more than twice its width and its perimeter is 300 meters, which of the following equations could be used to find L?

Aug 3, 2015

$L = \text{101 m}$
$w = \text{49 m}$

#### Explanation:

You have two equations to work with, the equation that describes the perimeter of the rectangle and the equation that describes the relationship between its length, $L$, and width, $w$.

The perimeter of a reactangle is always equal to

$P = 2 \cdot \left(L + w\right)$

$2 \left(L + w\right) = 300$

Now, you know that if you double the width and add three meters to the result, you get $L$. This means that

$L = 2 \cdot w + 3$

You now have a system of two equations

$\left\{\begin{matrix}2 L + 2 w = 300 \\ L = 2 w + 3\end{matrix}\right.$

Use the value of $L$ from the second equation to replace $L$ in the first equation. This will get you

$L = 2 w + 3$

$2 \cdot \left(2 w + 3\right) + 2 w = 300$

$4 w + 6 + 2 w = 300$

$6 w = 294 \implies w = \textcolor{g r e e n}{49}$

This means that $L$ is equal to

$L = 2 \cdot w + 3$

$L = 2 \cdot 49 + 3 = \textcolor{g r e e n}{101}$

The rectangle will thus be 49 m wide and 101 m long.