Question

The perimeter of a rectangle is 86 cm. twice the width exceeds the length by 2cm. write a system and solve to find the dimensions of the rectangle.

Answer 1

There are two equation of the system
(1) The perimeter
     2(w + l) = p
     2(w + l) = 86
(2) The length and the width
      2w = l + 2

From the second equation
2w = l + 2
l + 2 = 2w
l = 2w - 2

Subtitute l with 2w - 2 in the first equation to find the width
2(w + l) = 86
2(w + 2w - 2) = 86
2(3w - 2) = 86
6w - 4 = 86
6w = 90
w = 15

Now, find the length by subtituting w with 15
l = 2w - 2
l = 2(15) - 2
l = 30 - 2
l = 28

The dimension of the rectangle
length = 28 cm
width = 15 cm

Answer 2

Length = 28 cm and width = 15 cm

Let Length = x

width = y,

According to the question:

2(x + y) = 86 or, x + y = 43 or y = 43 - x.

2y = x + 2

Plug y = 43 - x in it.

$2y=x+2\\2(43-x)=x+2\\86-2x=x+2\\3x=84\\x=28$

So, y = 43 - 28 = 15

So, length = 28 cm and width = 15 cm

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