SOLUTION: The length of a rectangle is one foot more than twice the width. The perimeter is 20 feet. What is the width of the rectangle?
What is the solution? 4x-2(x-1)=12
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What is the solution? 4x-2(x-1)=12
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Question 261106: The length of a rectangle is one foot more than twice the width. The perimeter is 20 feet. What is the width of the rectangle?
What is the solution? 4x-2(x-1)=12
You can put this solution on YOUR website! The length of a rectangle is one foot more than twice the width. The perimeter is 20 feet. What is the width of the rectangle?
What is the solution? 4x-2(x-1)=12
Let the width be x
twice width = 2x
one more than twice the width = 2x+1 = length
Perimeter = 2*(l+w)
20= 2 * (x+2x+1)
20= 2x+4x+2
20=6x+2
20-2=6x
18/6=x
x=3 width
