Can you tell me how to figure this problem: The width of the rectangle is 5 ft more than twice the length.  If the area is 75 sq. ft., find the width.

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You can only solve for one variable at a time, so you have to make some changes to the equation do this. 1) The secret is in the first line. Turn that into "W=" form. The tricky part is to go past the words "more than". These words really mean "twice the length plus and extra 5ft." So you get W=(2L+5).2) Now look at the area formula "A= L xW". It has 3 different variables, but you can make some substitutions.~ You have the information that the area is 75 sq ft so for "A" put in 75. ~ for "W" put in the (2L+5) that you created.~ A= LXW becomes 75 = L( 2L+5) (only 1 variable!!!)3) Now you can solve for L.75 = 2L^2 +5L (expand)  0 = 2L^2+5L-75 (rearrange)4) This is a quadratic equation, so factor the trinomial or use the quadratic formula to find L.   0 = (L-5)(2L+15)  L = 5, -7.55) Lengths can't be negative so L=5 ft.6) Test the answer in your original equation to be sure it works.75 = L( 2L+5) --> 75 = 5(2x5 + 5). It works!