SOLUTION: The length of a rectangle is 3 less than twice its width. The area of the rectangle is 35 cm^2. Find the dimensions of the rectangle.

Question 953981: The length of a rectangle is 3 less than twice its width. The area of the rectangle is 35 cm^2. Find the dimensions of the rectangle.
Found 2 solutions by amarjeeth123, addingup:
Answer by amarjeeth123(570) (Show Source): You can put this solution on YOUR website!
Let the width be x cm.
Then the length is (2x-3) cm.
Area of the rectangle=length* width
x(2x-3)=35
2x^2-3x-35=0
2x^2-10x+7x-35=0
2x(x-5)+7(x-5)=0
(2x+7)(x-5)=0
(2x+7)=0 or (x-5)=0
x=-7/2 or x=5
The width of the rectangle is 5cm.
The length of the rectangle is 7cm.
Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
Let's call the length L and the width W.
The area is:
W*L = 35, and the problem says:
L = 2W-3 Let's plug this value for L into the first equation:
W(2W-3)= 35 Multiply on left:
2W^2-3W= 35 Subtract 35 from both sides:
2W^2-3W-35= 0 Now we have quadratic equation. Let's factor it:
We have a -3W in the middle. And we know we need two numbers, let's call them a and b, such that a+b = -3W And those same two numbers must: a*b = -35:
(W-5)(2W+7) = 0 Solve separately:
W-5 = 0 or 2W+7 = 0 On the first equation, add 5 to both sides. On the 2nd subtract 7 to both sides:
W= 5 or 2W= -7 On the second equation divide both sides by 2:
W= 5 or W= -7/2= -3.5
Let's discard the 2nd equation because we need a positive number. Let's take 5 and see. We said the area is:
W(2W-3)= 35
5(2*5-3)= 35 Solve to remove parenthesis:
5*7 = 35 We have the correct answer, the W=5 and the L= 7
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