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:Quote:
Originally Posted by DPXJube
2x + 2y = 86
2x - 2 = y
Solve the sytem.
Not to sure how you deprived this from 2x+2y=86Quote:
Originally Posted by undefined
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."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).Quote:
Originally Posted by DPXJube
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