# The width of a rectangle is 9 cm less than twice the length. The perimeter is 72 cm. What is the length and width of the rectanlge?Equation: Solution:

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The problem as stated is a two variable problem. That means there are two quantities in the problem which are unknown: the length and the width. To solve problems of this nature there must be two equations which relate the two unknown variables. In this case, we use the formulae for the perimeter and the given relationship between the length and the width.

`P = 2L + 2W`

and

`W = 2L-9`

Substitute the expression for the width into the equation for the perimeter:

`P = 2L + 2(2L-9)`

Expand the parenthesis by applying the distributive property and replace P with the given value for the perimeter:

`72 = 2L + 4L - 18`

Collect similar terms on the right side of the equation, add 18 to both sides, and apply the symmetric property of equality:

`72 = 6L - 18`

`6L = 90`

Divide both sides by 6 to determine that the length is 15.

Substitute this back into the width equation to determine W:

`W = 2x15 - 9 = 21`

Check your answer by computer the perimeter:

2x15 + 2x21 = 30 +42 = 72

State the answer using proper units of measure:

**L = 15 cm and W = 21 cm or "The length of the rectangle is 15 cm and the width is 21 cm"**

2*（X+2X-9）=72

3X-9=36

length: X=15

width:2X-9=21