# The area of a rectangle is 42 yd^2 , and the length of the rectangle is 11 yd less than three times the width, how do you find the dimensions length and width?

Feb 23, 2016

The dimensions are as follows:
Width$\left(x\right) = 6$ yards
Lenght $\left(3 x - 11\right) = 7$ yards

#### Explanation:

Area of rectangle $= 42$ square yards.

Let the width $= x$ yards.

The length is 11 yards less than thrice the width:

Length $= 3 x - 11$ yards.

Area of rectangle $=$ length $\times$ width

$42 = \left(3 x - 11\right) \times \left(x\right)$

$42 = 3 {x}^{2} - 11 x$

$3 {x}^{2} - 11 x - 42 = 0$

We can Split the Middle Term of this expression to factorise it and thereby find the solutions.

$3 {x}^{2} - 11 x - 42 = 3 {x}^{2} - 18 x + 7 x - 42$

$= 3 x \left(x - 6\right) + 7 \left(x - 6\right)$

$\left(3 x - 7\right) \left(x - 6\right)$ are the factors, which we equate to zero in order to obtain $x$

Solution 1:
$3 x - 7 = 0 , x = \frac{7}{3}$ yards (width).
Length $= 3 x - 11 = 3 \times \left(\frac{7}{3}\right) - 11 = - 4$ yards, this scenario is not applicable.

Solution 2:

$x - 6 = 0 , x = 6$ yards (width) .

Length $= 3 x - 11 = 3 \times 6 - 11 = 7$ yards (length).