# The area of a rectangle is 65yd^2 , and the length of the rectangle is 3yd less than double the width, how do you find the dimensions?

build the equations and solve...

#### Explanation:

let the area be $A = l \cdot w$ where length is $l$ and width is $w$

so 1.st equqtion will be

$l \cdot w = 65$

and length is 3 yd less than double the width says:

$l = 2 w - 3$ (2nd eq.)

replacing $l$ with $2 w - 3$ in first eq. will yield

$\left(2 w - 3\right) \cdot w = 65$
$2 {w}^{2} - 3 w = 65$
$2 {w}^{2} - 3 w - 65 = 0$

now we have a 2nd order equation just find the roots and take the positive one as width can not be negative...

$w = \frac{3 \pm \sqrt{9 + 4 \cdot 2 \cdot 65}}{2 \cdot 2} = \frac{3 \pm \sqrt{529}}{4} = \frac{3 \pm 23}{4}$
$w = - 5 , \frac{13}{2}$ so taking $w = \frac{13}{2} = 6.5 y d$

replacing $w$ with $6 , 5$ in second eq. we get

$l = 2 w - 3 = 2 \cdot 6.5 - 3 = 13 - 3 = 10 y d$

$A = l \cdot w = 10 \cdot 6.5 = 65 y d$ will confirm us...