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Thread: area word problem

  1. #1
    Junior Member
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    area word problem

    Q:if a rectangle were 3 in longer and 1 in narrower it would contain 5 sq.in more than it does now; but if it were 2 in shorter and 2 in wider its area would remain unchanged. What are dimensions?

    I understand that the question leads to a simultaneous eq.
    it says it would contain 5 sq in
    does it mean that it equals 5
    therefore $(x+3)(y-1) =5$
    i formed this simultaneous eq

    $(x+3)(y-1) =5$
    $(x-2)(y+2) =5$
    but it was wrong

    I think i didnt interrupt the question probably
    please tel me where i am going wrong

    thank ou
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  2. #2
    MHF Contributor
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    Re: area word problem

    Quote Originally Posted by bigmansouf View Post
    Q:if a rectangle were 3 in longer and 1 in narrower it would contain 5 sq.in more than it does now; but if it were 2 in shorter and 2 in wider its area would remain unchanged. What are dimensions?

    I understand that the question leads to a simultaneous eq.
    it says it would contain 5 sq in
    does it mean that it equals 5
    therefore $(x+3)(y-1) =5$
    i formed this simultaneous eq

    $(x+3)(y-1) =5$
    $(x-2)(y+2) =5$
    but it was wrong

    I think i didnt interrupt the question probably
    please tel me where i am going wrong

    thank ou
    you want

    $(x+3)(y-1) =x y + 5$
    $(x-2)(y+2) =x y$
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  3. #3
    MHF Contributor

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    Re: area word problem

    Quote Originally Posted by bigmansouf View Post
    Q:if a rectangle were 3 in longer and 1 in narrower it would contain 5 sq.in more than it does now; but if it were 2 in shorter and 2 in wider its area would remain unchanged. What are dimensions?

    I understand that the question leads to a simultaneous eq.
    it says it would contain 5 sq in
    does it mean that it equals 5
    The problem doesn't say "it would contain 5 sq in". It says "it would contain 5 sq in more".
    With sides x and y, in inches, the original area was xy sq. in. "5 sq in more" would be xy+ 5.

    therefore $(x+3)(y-1) =5$
    i formed this simultaneous eq

    $(x+3)(y-1) =5$
    $(x-2)(y+2) =5$
    but it was wrong

    I think i didnt interrupt the question probably
    please tel me where i am going wrong

    thank ou
    Follow Math Help Forum on Facebook and Google+

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