Three times the width of a certain rectangle exceeds twice its length by three inches, and four times its length is twelve more than its perimeter. Find the dimensions of the rectangle.
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Algebraic Steps

The first statement compares the length L and the width W. Start by doing things orderly, with clear and complete labelling:
three times the width: 3W
twice its length: 2L
exceeds by three inches, meaning “is three inches greater than”: + 3
equation: 3W = 2L + 3
Now I have the second statement, which compares the length L and the perimeter P. I will be complete with my labelling:
four times its length: 4L
perimeter: P = 2L + 2W (this is just the perimeter formula for rectangles)
twelve more than: + 12
equation: 4L = P + 12, or 4L = (2L + 2W) + 12 (by substitution)
3W = 2L + 3
4L = 2L + 2W + 12
There are various ways of solving this; the way I do it (below) just happens to be what I thought of first. I’ll take the first equation and solve for W:
3W = 2L + 3
W = ( 2/3 )L + 1
Now I’ll simplify the second equation, and then plug in this above expression for W:
4L = 2L + 2W + 12
2L = 2W + 12
2L = 2[ ( 2/3 )L + 1 ] + 12 (by substitution from above)
2L = ( 4/3 )L + 2 + 12
2L = ( 4/3 )L + 14
2L – ( 4/3 )L = 14
( 6/3 )L – ( 4/3 )L = 14
( 2/3 )L = 14
L = (14)×( 3/2 ) = 21
Then:
W = ( 2/3 )L + 1
= ( 2/3 )×(21) + 1
= 14 + 1 = 15
Now, remember that the question didn’t ask “Find the values of the variables L and W”. It asked you to “Find the dimensions of the rectangle,” so the actual answer is:
The length is 21 inches and the width is 15 inches.
(Always be sure to check for the appropriate units, which was “inches” in this case.)
Some problems are just straightforward applications of basic geometric formulae.