# The length of a rectangle is 3 yard less than twice the width, and the area of the rectangle is 44 yd^2. How do you find the dimensions of the rectangle?

Mar 15, 2016

Length = $8$ yards
Width = $\frac{11}{2} = 5 \frac{1}{2}$ yards

#### Explanation:

Let
color(white)("XXX")L: " length of rectangle (in yards)"
color(white)("XXX")W: " width of rectangle (in yards)"

We ar told
[1]$\textcolor{w h i t e}{\text{XXX}} L = 2 W - 3$
and
[2]$\textcolor{w h i t e}{\text{XXX}} L W = 44$

Substituting the expression $\left(2 W - 3\right)$ from [1] for $L$ in [2]
[3]$\textcolor{w h i t e}{\text{XXX}} \left(2 W - 3\right) W = 44$

[4]$\textcolor{w h i t e}{\text{XXX}} 2 {W}^{2} - 3 W - 44 = 0$

We could try to factor this (it can be done)
$\textcolor{w h i t e}{\text{XXX}} W = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(2\right) \left(- 44\right)}}{2 \left(2\right)}$
$\textcolor{w h i t e}{\text{XXXX}} = \frac{3 \pm \sqrt{361}}{4}$
$\textcolor{w h i t e}{\text{XXXX}} = \frac{3 \pm 19}{4}$
Giving $W = \frac{22}{4} = \frac{11}{2} \rightarrow L = 8$ (using [2])
The other algebraic solution: $W = \frac{- 16}{4}$ is obviously extraneous since the width can not be nedgative.