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Question

the perimeter of a rectangle is 240 inches. the length excees the width by 60 inches. the the length and the width

Submitted: 89 days ago.

Category: Math

Value: $9

Status: CLOSED

Level/Year: sophmore college
Subject: college algebra

Already Tried:
trying to write two differnt equations and solve for both.

Accepted Answer

Let width = x, then length = x + 60

Perimeter = 2(length + width)

240 = 2(x + 60 + x)

240 = 2(2x + 60)

120 = 2x + 60

2x = 60

x = 30 inches (width) and 30 + 60 = 90 inches (length).

Accepts:
Expert:	Chirag
Pos. Feedback:	99.3 %
Answered:	8/25/2009

Tutor

Masters Degree

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Math

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Question

the perimeter of a rectangle is 240 inches. the length excees the width by 60 inches. the the length and the width

Optional Information

Level/Year: sophmore college
Subject: college algebra

Already Tried:
trying to write two differnt equations and solve for both.

Accepted Answer

Let width = x, then length = x + 60

Perimeter = 2(length + width)

240 = 2(x + 60 + x)

240 = 2(2x + 60)

120 = 2x + 60

2x = 60

x = 30 inches (width) and 30 + 60 = 90 inches (length).

Math

Have your own Math question?

Related Math Topics:

Question

the perimeter of a rectangle is 240 inches. the length excees the width by 60 inches. the the length and the width

Optional Information

Level/Year: sophmore collegeSubject: college algebra Already Tried: trying to write two differnt equations and solve for both.

Accepted Answer

Let width = x, then length = x + 60 Perimeter = 2(length + width) 240 = 2(x + 60 + x) 240 = 2(2x + 60) 120 = 2x + 60 2x = 60 x = 30 inches (width) and 30 + 60 = 90 inches (length).

Related Math Questions

Level/Year: sophmore college
Subject: college algebra

Already Tried:
trying to write two differnt equations and solve for both.

Let width = x, then length = x + 60

Perimeter = 2(length + width)

240 = 2(x + 60 + x)

240 = 2(2x + 60)

120 = 2x + 60

2x = 60

x = 30 inches (width) and 30 + 60 = 90 inches (length).