SOLUTION: I think this is a linear equation, but anyways, here's the problem:
The length of a room is four times its width. If the perimeter is 250 feet, what are the dimensions of the ro
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-> SOLUTION: I think this is a linear equation, but anyways, here's the problem:
The length of a room is four times its width. If the perimeter is 250 feet, what are the dimensions of the ro
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Question 1013409: I think this is a linear equation, but anyways, here's the problem:
The length of a room is four times its width. If the perimeter is 250 feet, what are the dimensions of the room? I set up two equations:
1) 4l +w=250
2) l -w =0
I used the method of elimination and ended up with:
5l = 250 and then I divided by 5.
The result was l = 50. I divided that by 4 and got 12.5 inches which I presumed was the measurement of the width. My final answer was:
l=50 in and w=12.5 in (there is no dimension specified, but I used inches)
Am I correct, or was I supposed to plug 50 back in to the original equation?
Thanks for helping, and please show your work! Answer by josgarithmetic(39617) (Show Source):
You can put this solution on YOUR website! The character which looks like the ONE digit is a very bad choice for a variable because it interferes with reading.
L for length, w for width.
The length of a room is four times its width, ;
(