SOLUTION: Hello, I have a problem that has been answered here already. The problem is, I don't quite understand how he arrived at the answer. The problem and answer is here: http://www.algeb

Question 51177: Hello, I have a problem that has been answered here already. The problem is, I don't quite understand how he arrived at the answer. The problem and answer is here: http://www.algebra.com/cgi-bin/jump-to-question.mpl?question=19628
The original question is: Suppose that the length of a rectangle is one and one-third times as long as its width. The area of the rectangle is 48 square centimeters. Find the length and width of the rectangle.
The answer given is:
LET WIDTH =W
LENGTH=one and one-third times as long as its width.=(1 and 1/3)(W)..=(4/3)(W).
=4W/3
SO AREA =L*W=(4W/3)*W=48
W*W=48*3/4=36
W=6 CM
L=4W/3=4*6/3=8 CM
I understand most of it, until the line w*w=48*3/4=36 , I'm not sure I understand how he turned 4/3 into 3/4 and multiplied it by 48. I thought one could only add or subtract when bringing a fraction across? Sorry if this is a dumb question.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
(4W/3)*W=48
(4/3)W*W = 48
Now multiply both sides by (3/4) which is the reciprocal of (4/3)
to get;
(3/4)(4/3)W*W = 48*(3/4)
Do you see that (3/4)(4/3)= 1 ?
W*W=48*(3/4)
W*W=36
W=6 CM
L=4W/3=4*6/3=8 CM
Cheers,
Stan H.
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