Question 228533
Solve algebraically using only one variable: the length of a rectangle is two more than twice its width. If the area of the rectangle is 84, find the length and the width. 

Step 1.  Area A of a rectangle is A=width times the length


Step 2.  Let w be the width.


Step 3.  Let 2w+2 be the length since the length it's two more than twice its width.


Step 4.  Then, {{{A=w*(2w+2)=2w^2+2w=84}}}. 


Step 5.  Solving yields the following steps:


Subtract 84 from both sides of the equation


{{{2w^2+2w-84=84-84}}}


{{{2w^2+2w-84=0}}}


Step 6. To solve, use the quadratic equation given as 


{{{w = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


where a=2, b=2, and c=-84.


*[invoke quadratic "w", 2, 2, -84 ]


Selecting the positive solution


{{{w=6}}} and {{{2w+2=14}}} and the A=6*14=84 which is a true statement.


Step 6.  ANSWER:  The width is 6 and the length is 14.


I hope the above steps were helpful.


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Good luck in your studies!


Respectfully,
Dr J