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

$\setminus \textrm{L e n > h} = 10$, $\setminus \textrm{w i \mathrm{dt} h} = \frac{13}{2}$

#### Explanation:

Let $L$ & $B$ be the length and width of rectangle then

as per given condition

$L = 2 B - 3 \setminus \ldots \ldots \ldots . \left(1\right)$

And the area of rectangle

$L B = 65$

setting value of $L = 2 B - 3$ from (1) in above equation, we get

$\left(2 B - 3\right) B = 65$

$2 {B}^{2} - 3 B - 65 = 0$

$2 {B}^{2} - 13 B + 10 B - 65 = 0$

$B \left(2 B - 13\right) + 5 \left(2 B - 13\right) = 0$

$\left(2 B - 13\right) \left(B + 5\right) = 0$

$2 B - 13 = 0 \setminus \setminus \mathmr{and} \setminus \setminus B + 5 = 0$

$B = \frac{13}{2} \setminus \setminus \mathmr{and} \setminus \setminus B = - 5$

But the width of rectangle can't be negative hence

$B = \frac{13}{2}$

setting $B = \frac{13}{2}$ in (1), we get

$L = 2 B - 3$

$= 2 \left(\frac{13}{2}\right) - 3$

$= 10$