Question

Help sorry nothing was attached last time

Answer 1

Given:

The area of a rectangle is 99 square yd.

Length of the rectangle = 7 yd more than twice the width.

To find:

The dimensions of the rectangle.

Solution:

Let x be the width of the rectangle. Then, length of the rectangle is:

$Length=2x+7$

Area of a rectangle is:

$A=length\times width$

$A=(2x+7)\times x$

$A=2x^2+7x$

The area of a rectangle is 99 square yd.

$2x^2+7x=99$

$2x^2+7x-99=0$

Splitting the middle term, we get

$2x^2+18x-11x-99=0$

$2x(x+9)-11(x-9)=0$

$(x+9)(2x-11)=0$

Using zero product property, we get

$(x+9)=0$ and $(2x-11)=0$

$x=-9$ and $x=\dfrac{11}{2}$

$x=-9$ and $x=5.5$

Width of the rectangle cannot be negative. So, $x=5.5$ yd.

Now, the length of the rectangle is:

$Length=2x+7$

$Length=2(5.5)+7$

$Length=11+7$

$Length=18$

Therefore, the length of the rectangle is 18 yd and the width of the rectangle is 5.5 yd.