# Finding Perimeter of a rectangle when not given any side length

• Jun 20th 2010, 02:42 PM
DPXJube
Finding Perimeter of a rectangle when not given any side length
The question in particular is that I am given a rectangle where the length is 2 m less then twice the width. The perimeter is 86 m and I need to find the side length.

I am aware the the answer is that the length is 28 and the width is 15, but I need to show this algebraically. Help would be much obliged.
• Jun 20th 2010, 02:45 PM
undefined
Quote:

Originally Posted by DPXJube
The question in particular is that I am given a rectangle where the length is 2 m less then twice the width. The perimeter is 86 m and I need to find the side length.

I am aware the the answer is that the length is 28 and the width is 15, but I need to show this algebraically. Help would be much obliged.

Let width be x and length be y. We set up a system of linear equations:

2x + 2y = 86
2x - 2 = y

Solve the sytem.
• Jun 20th 2010, 02:51 PM
DPXJube
Quote:

Originally Posted by undefined
2x - 2 = y

Not to sure how you deprived this from 2x+2y=86
• Jun 20th 2010, 02:55 PM
undefined
Quote:

Originally Posted by DPXJube
Not to sure how you deprived this from 2x+2y=86

The second equation is not derived from the first. It is derived from the problem statement: "the length is 2 m less then twice the width."
• Jun 20th 2010, 02:59 PM
DPXJube
Okay. So then how do I solve for x?
• Jun 20th 2010, 03:12 PM
undefined
Quote:

Originally Posted by DPXJube
Okay. So then how do I solve for x?

This is a question about solving a system of (linear) equations, and the techniques are: elimination (multiply equations by nonzero constants and add them) and substitution (solve for a variable and substitute into another equation).

So example of elimination: subtract the second equation from the first to get

2y+2 = 86-y

y = 28

Example of using substitution:

2x + 2(2x - 2) = 86

6x = 90

x = 15