The length of a box is four-and-a-half inches more than its height. The width of the box is seven inches more than it's height. If the volume of the box is 4981.5 cubic inches, then what is the height of the box?

The height of the box is 13.5 inches.

Step-by-step explanation:

Assume that the height of the box be x inches.

The length of the box four and half inches more than the height.

So, the height of the box is

inches

The width of the box 7 inches more than its height

The width of the box is =(x+7) inches

So, the volume of the box is =Length×width×height

                                              cubic inches

                                             ={x(x+7)+4.5(x+7)}x cubic inches

                                             ={x²+7x+4.5x+31.5}x cubic inches

                                             =(x³+11.5x²+31.5x)  cubic inches

But give that, the volume of the box is 4981.5 cubic inches.

According to the problem,

x³+11.5x²+31.5x=4981.5

⇒x³+11.5x²+31.5x-4981.5=0

⇒x³-13.5x²+25x²-337.5x+369x-4981.5=0

⇒x²(x-13.5)+25x(x-13.5)+369(x-13.5)=0

⇒(x-13.5)(x²+25x+369)=0

∴x=13.5 inches

The height of the box is 13.5 inches.

but which class are u may know that first

use distributive property on all like terms and change your signs if you have a double negative or if you have positive/negative