Question 302686
The perimeter of a rectangle is 20 meters.
 If the length is increased by four times the width, the sum is 19 meters.
 What are the dimensions of the rectangle? What is its area?
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Let L = the length of the rectangle
Let W = the width
:
Write an equation for each statement
"The perimeter of a rectangle is 20 meters."
2L + 2W = 20
We can simplify this divide by 2, results:
L + W = 10
W = (10-L); this form for substitution
:
"the length is increased by four times the width, the sum is 19 meters."
L + 4W = 19
Substitute W with (10-L)
L + 4(10-L) = 19
L + 40 - 4L = 19
L - 4L = 19 - 40
-3L = - 21
L = {{{(-21)/(-3)}}}
L = +7m is the length
:
What are the dimensions of the rectangle?
find the width
W = 10 - L
W = 10 - 7
W = 3m
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Dimensions: 7 by 3
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What is its area? 7 * 3 = 21 sq/m
:
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Check solutions by finding the perimeter with these dimensions
2(7) + 2(3) = 20