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The perimeter of a rectangle is 120 meters, and the length is 20 meters longer than the width. Find the dimensions of the rectangle.

Let \( x \) represent the length and \( y \) represent the width. The corresponding system of equations is:

1. \( 2x + 2y = 120 \)
2. \( x - y = 20 \)

Answer :

Final answer:

The dimensions of the rectangle are length = 33 meters and width = 13 meters.

Explanation:

To find the dimensions of the rectangle, we can set up a system of equations based on the given information. Let x be the length and y be the width of the rectangle.

From the problem, we have the following equations:

  1. 2x + 2y = 120 (equation for the perimeter)
  2. x - y = 20 (equation for the length being 20 meters longer than the width)

We can solve this system of equations by using substitution or elimination. Let's use substitution method:

From equation 2, we can solve for x:

x = y + 20

Substitute this expression for x in equation 1:

2(y + 20) + 2y = 120

Simplify the equation:

4y + 40 + 2y = 120

Combine like terms:

6y + 40 = 120

Subtract 40 from both sides:

6y = 80

Divide both sides by 6:

y = 80/6

Simplify the fraction:

y = 13.33

Round y to the nearest whole number: y ≈ 13

Substitute this value for y in the expression for x:

x = 13 + 20

x = 33

So, the dimensions of the rectangle are: length = 33 meters and width = 13 meters.

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Rewritten by : Barada