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* width`

`A=(2x+7)* 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=(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.