algebra1
posted by anonymous
The length of a rectangle is twice the width. If the length is increased by 4 inches and the width is decreased by 1 inch, a new rectangle is formed whose perimeter is 198 inches. Find the dimensions of the original rectangle.

Reiny
just translate the "English" into Math
"The length of a rectangle is twice the width" > L = 2w
"If the length is increased by 4 inches" > 2w + 4
"the width is decreased by 1 inch" > w+1
"perimeter is 198 inches" > 2(2w+4) + 2(w+1) = 198
Solve for w, let me know what you got 
Reiny
typo error, should have been
"the width is decreased by 1 inch" > w1
"perimeter is 198 inches" > 2(2w+4) + 2(w1) = 198 
anonymous
i got x= 32

Reiny
correct, so the original width was 32 in and the length was 64 inches.

anonymous
That is how far I got on my paper, but my teacher says I need to find the new length and width as well. How would I go about doing that?

Reiny
well, didn't we call the new length 2w+4, put in w=32 to get 68
and didn't we call the new width w1, put in w=32 into that to get 31
check: new perimiter = 2(68)+2(31)
=136+62
=198 !!!! 
anonymous
Oh,thank you so much so much for your time, i understand now!=)
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