Question

The perimeter of a rectangle is 42 inches, and its area is 110 square inches. find the length and width of the rectangle.

Answer 1

`p = 2w + 2l`
`p = 42`

`a = l * w`
`a = 108`

You want to make either the length or width variable (I chose the width variable to be by itself. It doesn't matter) by itself by:

Divide both sides by 2.
(p) ÷ 2 = (2w + 2l) ÷ 2

`2's`
`cancel`
`out`

`(p)/(2) = w + l`

Subtract both sides by either length or width depending on what variable you chose to make by itself (in this case I subtracted the length variable because I wanted the width variable by itself).

`(p)/(2) - l = w + l - l`

`l's`
`cancel`
`out`
p/2 - l = w


`a = l * w`

You replace w with the equation above p/2 - l = w

`a = l( (p)/(2) - l)`

Multiply out the L.

`a = (p)/(2)l - l^(2)`

Plug in all your numbers a = 110 and p = 42

`110 = (42)/(2)l - l^(2)`

`110 = 21l - l^{2`

Move everything to the left side so
`l^(2)` would be positive (makes the equation easier when
`l^(2)` is positive).

`l^(2) - 21l + 110 = 0`

Factor.

`(l -10)(l-11) = 0`

Make each parenthesis set equal to 0.

`l-11=0`
`l-10=0`

Add.

`l = 11`
`l = 10`

By doing this you solve for both the width and length so the answer is:
w = 11 L = 10