the perimeter of a rectangle is 44cm. the length is 7 more then twice the width. what is the length/width?

the perimeter of a rectangle is 44cm. the length is 7 more then twice the width. what is the length/width?

width of the rectangle: W

length of the rectangle: L

the perimeter: $$p = 2 * ( L+W ) = 44\;cm \quad \mbox{or} \quad \boxed{L + W = 22 \;cm \qquad (1)}$$

the length is 7 more then twice the width: $$L = 7\;cm + 2W \quad \mbox{or} \quad \boxed{L - 2W = 7\;cm \qquad (2)}$$  

  $$\begin{array}{lrcl} (1) - (2):&\\ &\underbrace{L-L}_{=0}+\underbrace{W-(-2W)}_{=3W}&=&\underbrace{22 cm - 7cm}_{=15\;cm}\\ &3W&=&15\;cm \quad | \quad : 3\\ &W&=&5\;cm \end{array}$$

$$\begin{array}{rcl} L + W &=& 22\;cm\\ L &=& 22 \;cm- W \\ L&=& 22\;cm- 5 \;cm\\ L &=& 17 \;cm \end{array}$$

Let's set up an equation after defining some variables:

L = length & W = width

L = 2W + 7 (length is 7 more than twice the width)

Perimeter of a rectangle is 2L + 2W...

2(2W + 7) + 2W = 44.

Distribute the 2.

4W + 14 + 2W = 44.

Combine like terms and subtract 14 from both sides.

6W = 30.

Solve for W.

W = 30/6 = 5.

So, L = 2W + 7 = 2(5) + 7 = 10 + 7 = 17.

L = 17 cm.

W = 5 cm.

Check: 2L + 2W = 44. 

2(5) + 2(17) = 44.

10 + 34 = 44. Yes.