Help with perimeter math problem? (9th grade algebra)?

The perimeter of a rectangle is 42 centimeters. The length of the rectangle can be represented by (x+4), and its width can be represented by (2x-7). What are the dimensions of this rectangle in centimeters?

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  • The perimeter of a rectangle is 42 centimeters. The length of the rectangle can be represented by (x+4), and its width can be represented by (2x-7). What are the dimensions of this rectangle in centimeters?

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    perimeter of a rectangle is 2(L+W); this means L+W = 21.

    so (x+4) +(2x-7) = 21

    expand and combine like terms

    3x-3 = 21

    add 3 to both sides

    3x = 24; x = 8

    L = x+4, or 12 cm

    W = 2x-7 or 9 cm

  • Perimeter of a rectangle = 2(length + width)

    length = (x + 4)

    width = (2x - 7)

    P = 2[(x + 4) + (2x - 7)] = 6x - 6 = 42

    x = 8

    the length is: 12 cm

    the width is: 9 cm

    check...I did

    archimedes

  • Perimeter = 2 ( length + width )

    42 = 2 ( x+4 + 2x -7)

    21 = 3x -3

    3x = 24

    x = 8

    Length = 12 cm Width = 9 cm ANSWER

  • 2(x+4) + 2(2x-7) = 42

    2x + 8 + 4x - 14 = 42

    6x - 6 = 42

    6x = 48

    x = 8

    dimensions

    length: x+4 = 8+4 = 12 cm

    width: 2x-7 = 2*8 -7 = 9 cm

  • so to set this up you want to do perimeter= (length*2)+(width*2)

    so now add in your given info 42=((x+4)*2)+((2x-7)*2)

    first simplify the times two into the x+4 and 2x-7 to get 42=2x+8+4x-14

    now combine like terms 42=6x-6

    add the 6 onto the 42 to cancel it from 6x to get 48=6x

    get x by itself by doing 48/6 to get 8=x

    now plug 8 in for x on your length and width so now your Length should be (8+4) which is 12

    and 2(8)-7 which is 9

    if you want to test this do(12*2)+(9*2)

    so your answer is 12cm and 9 cm

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