SOLUTION: If the Length of a rectangle is five less than twice width, and the perimeter is 86 feet, find the dimensions.
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Question 882951: If the Length of a rectangle is five less than twice width, and the perimeter is 86 feet, find the dimensions.
Answer by Leaf W.(135) (Show Source): You can put this solution on YOUR website!
Hello!
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Let us call the width of the rectangle "x." Since the length of the rectangle is five less than twice its width," we can express the length of the rectangle as 2x - 5 (since 2x = twice the width and - 5 indicates "five less").
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Now, using x as the width and 2x - 5 as the length, we can create an equation. We know that the perimeter is 86 feet, and the perimeter of a rectangle is the same thing as adding up its sides together: length + length + width + width, or 2*length + 2*width. Thus, we can create the equation
2(length) + 2(width) = 86
Substituting in our expressions for width and length, we get
2(2x - 5) + 2(x) = 86
Now we can just solve for x
2(2x - 5) + 2x = 86
Distribute: 4x - 10 + 2x = 86
Add like terms: 6x - 10 = 86
Add 10 to both sides: 6x = 96
Divide both sides by 6: x = 16
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Therefore, x, which represents the width of the rectangle, is 16 feet. The length must then be 2x - 5 = 2(16) - 5 = 32 - 5 = 27 feet.
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Thus, the length of the rectangle is 27 feet, and the width is 16 feet.
Let me know if you need further clarification! =)
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