algebra1

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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.

• algebra1 -

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

• algebra1 -

typo error, should have been

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

"perimeter is 198 inches" ----> 2(2w+4) + 2(w-1) = 198

• algebra1 -

i got x= 32

• algebra1 -

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

• algebra1 -

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?

• algebra1 -

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 !!!!

• algebra1 -

Oh,thank you so much so much for your time, i understand now!=)