Question

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.
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
typo error, should have been

<b>"the width is decreased by 1 inch" ---> w-1

"perimeter is 198 inches" ----> 2(2w+4) + 2(w-1) = 198 </b>
i got x= 32
correct, so the original width was 32 in and the length was 64 inches.
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?
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 w-1, put in w=32 into that to get 31

check: new perimiter = 2(68)+2(31)
=136+62
=198 !!!!
Oh,thank you so much so much for your time, i understand now!=)

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