Question

What would be the equation (using x) to solve “What are the dimensions of a rectangle whose length is 4 more than twice the width and whose perimeter is 3 less than 7 times the width?”

Answers

  1. Let w = the width
    Let 2w + 4 = the length
    Let 7w-3 = the perimeter
    P = 2l + 2w
    7w-3 = 2w + 2(2w+4)
    7w-3 = 2w + 4w + 8
    w = 11
    length = 2w + 4 = 2(11) + 4
    length = 26
    P = 7(11) -3 = 77 – 3
    Perimeter = 74

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  2. Answer:Let w = the width
    Let 2w + 4 = the length
    Let 7w-3 = the perimeter
    P = 2l + 2w
    7w-3 = 2w + 2(2w+4)
    7w-3 = 2w + 4w + 8
    w = 11
    length = 2w + 4 = 2(11) + 4
    length = 26
    P = 7(11) -3 = 77 – 3
    Perimeter = 74
    Step-by-step explanation:

    Reply

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