The length of a rectangle is 4 in. more than twice its width?

Question

If the perimeter of the rectangle is 56 in., find the width of the rectangle

Answer 1

Here is your system of equations:
L = 4 + 2w
P = 56 = 2L + 2w

Substitute first equation into the second:
56 = 2(4 + 2w) + 2w
56 = 8 + 4w + 2w
48 = 6w
w = 8

Answer 2

PERIMETER OF A RECTANGLE = 2(l+b)
let us suppose that the width is b
then ur length is 4+2b
now as u said perimeter = 56
so 2(4+2b+b)=56
or, 2(4+3b) =56
or, 4+3b =56/2
or, 3b = 28 - 4
or, b = 24/3 = 8
so l = 4 + 2*8 = 20

Answer 3

Width = x
Length = 4 + 2x

Perimeter = 2(length + width)
56 = 2(4+2x + x)
56 = 2(4 + 3x)
28 = 4 + 3x
24 = 3x
8 = x

Width = x = 8
Length = 4 + 2x = 4 + 2(8) = 20

Answer 4

let the width of the rectangle be = x in.
then its length = 2x + 4 in.
its perimeter = 2[ x + 2x + 4] in
thus we have:
2[x + 2x + 4] = 56
x + 2x + 4 = 28
3x = 24
x = 8

then width of the rectangle = 8 in.

check:
width = 8 in.
length = 16 + 4 = 20 in.
perimeter = 2(width + length) = 2(20+8) = 56 in.

Answer 5

Perimeter Formula

P = 2L + 2W

56 = 2(2w + 4) + 2w

56 = 4w + 8 + 2w

56 = 6w + 8

56 - 8 = 6w + 8 - 8

48 = 6w

48 / 6 = 6w / 6

48/6 = w

8 = w

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The width is 8 inches

The length = 2w + 4

2(8) + 4 =

16 + 4 =

20

The Length is 20 inches

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Answer 6

given L = 4 + 2W
perimeter = 2L + 2W = 56

substitute L,
2(4 + 2W) + 2W = 56
8 + 4W + 2W = 56
6W = 56 - 8 = 48
W = 48/6 = 8 in.

Answer 7

Let W = width and L = length
2L + 2W = 56
4W + 8 + 2W = 56
6W = 48
W = 8 in
L = 20 in