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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!=)

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