SOLUTION: please help me solve this. the length of a rectangle is seven units more then its width. If the width is doubled and the length is increased by two, the area is increased by for

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Question 844265: please help me solve this.
the length of a rectangle is seven units more then its width. If the width is doubled and the length is increased by two, the area is increased by forty-two square units. Find the dimensions of the original rectangle.
I have tried two rectangles one with x+7 by x with the area of x squared. Then a second rectangle with x+9 by 2x with the area of x squared+ forty-two. after that I get the second rectangle and set the formula of x2+42= 2x2+18x. I subtract the x2+42 to get the quadratic of x2+18x-42=0
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 

Hi,

Question states***

x(x+7) + 42 = 2x(x+9)

 x^2 + 7x + 42 = 2x^2 + 18x

    x^2 + 11x - 42 = 0

(x+14)(x-3) = 0   (Tossing out the negative solution for unit measure)

x = 3units, the width of the original rectangle.  the length is 10units

CHECKING our answer***



Wish You the Best in your Studies.



 

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