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