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 
               $2's$  $cancel$  $out$
$\frac{p}{2} = w + l$

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).
$\frac{p}{2} - l = w + l - l$
                 $l's$  $cancel$  $out$ 
p/2 - l = w

$a = l * w$

You replace w with the equation above p/2 - l = w 
$a = l( \frac{p}{2} - l)$

Multiply out the L.
$a = \frac{p}{2}l - l^{2}$

Plug in all your numbers a = 110 and p = 42
$110 = \frac{42}{2}l - l^{2}$
$110 = 21l - l^{2$

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