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

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> 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      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

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) About Me  (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***

3%2A10+%2B+42+=+6%2A12=+72+u%5E2

Wish You the Best in your Studies.