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The length of a rectangle is 2 feet less than three times the width. If the perimeter is 68 feet, what are the dimensions of the rectangle?

Answer :

The length of the rectangle is 25 feet and the width of the rectangle is 9 feet.

It is given that,

  • Length of a rectangle is 2 feet less than three times the width.
  • Perimeter is 68 feet.

Explanation:

Let [tex]x[/tex] be the width of the rectangle. Then the length of the rectangle is [tex]3x-2[/tex].

Perimeter of a rectangle is:

[tex]P=2(\text{length}+\text{width})[/tex]

[tex]68=2(3x-2+x)[/tex]

[tex]68=2(4x-2)[/tex]

[tex]68=8x-4[/tex]

Isolate the variable.

[tex]68+4=8x[/tex]

[tex]\dfrac{72}{8}=x[/tex]

[tex]9=x[/tex]

Now, the length of the rectangle is:

[tex]3(9)-2=27-2[/tex]

[tex]3(9)-2=25[/tex]

Thus, the length of the rectangle is 25 feet and the width of the rectangle is 9 feet.

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