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The length of a rectangle is nine more than twice the width. If the perimeter is 120 inches, find the dimensions.

The length of a rectangle is nine more than twice the width If the perimeter is 120 inches find the dimensions

Answer :

Answer:

Width = 17 inches

Length = 43 inches

Step-by-step explanation:

Framing and solving algebraic equation:

Let the width of the rectangle = x inches

Length = 2*width +9

= (2x + 9) cm

[tex]\boxed{\bf Perimeter \ of \ rectangle = 2* (length + width)}[/tex]

2* (length + width) = 120 inches

2* (2x + 9 + x) = 120

2* (3x + 9) = 120

Use Distributive property,

2 * 3x + 2*9 = 120

6x + 18 = 120

Subtract 18 from both sides,

6x = 120 - 18

6x = 102

Divide both sides by 6,

x = 102 ÷ 6

x = 17

Width = 17 inches

Length = 2*17 +9

= 34 + 9

= 43 inches

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