SOLUTION: The area of some rectangle is 35in^2. Four times the width of this rectangle is the same as 3 inches more than twice the length. What are the dimensions of the rectangle? I came

Question 913054: The area of some rectangle is 35in^2. Four times the width of this rectangle is the same as 3 inches more than twice the length. What are the dimensions of the rectangle?
I came up with the equations:
4W=2L+3 and LxW=35
I tried to do emilination, but because one of the equations is multiplying, it did not work. Should I substitute 4W=2L+3 with 4(35/L)=2L+3 and end up with a quadratic function & figure out the problem from there? Thank you so much for your time.
Answer by ichigo449(30) (Show Source): You can put this solution on YOUR website!
Yes that works. To continue: 4(35/L) = 2L+3 or L/2 + 3/4 = 35/L or (L^2)/2 + 3L/4 - 35 = 0 which is solvable by quadratic formula. Have a nice day.
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