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firefighter61 firefighter61
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11 years ago
The difference between the length and width of a rectangle is 2 units. The perimeter is 40 units. Write and solve a system of equations to determine the length and width of the rectangle.
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#1
11 years ago
L - W = 2
2W + 2L = 40

Solve for L in the first equation
L = W + 2

Plug it into the second equation
2W + 2(W + 2) = 40
4W  + 4  = 40
4W = 36
W = 9

Plug that into either equation to find length

L - 9 = 2
L = 11

Length is 11 units and width is 9 units Leftwards Arrow----ANSWER
wrote...
#2
11 years ago
l - w = 2
l + w = 20
Add together to get l= 11units and so
w = 9 units
wrote...
#3
11 years ago
l-w=2    so    l=w+2
2(l+w) =40

2l + 2w =40
2(w+2) + 2w =40
2w+4 +2w    =40
4w=36
w=9
l=11
wrote...
#4
11 years ago
length(L) - width(W) = 2
length(L) + width(W) = 20

Width (W)=9
Length (L)=11
wrote...
#5
11 years ago
2L+2W=40
W=L+2

2L+2W=40
2L+2(L+2)=40
2L+2L+4=40
4L+4-4=40-4
4L=36
L=9

W=L+2
W=9+2
W=11

The rectangle is 9 by 11
9+9+11+11 = 40
wrote...
#6
11 years ago
Let x be the length of this rectangle
and y be the width
so we got x-y=2
(x+y)*2=40

solving
x-y=2 or x=y+2 plug this value into (x+y)*2=40 we got
(y+2+y)*2=40 or (2y+2)*2=40
4y+4=40,  4y=36, so y=9
x=y+2=9+2=11

So here you have it, the length of this rectangle is 11 and the width is 9.
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