SOLUTION: The lenght of a rectangle is 3 cm greater than the width. If each dimension is increased by 2cm, the area is increased by 26 cm squared. Find the orginal dimensions of the rectan

Algebra ->  Surface-area -> SOLUTION: The lenght of a rectangle is 3 cm greater than the width. If each dimension is increased by 2cm, the area is increased by 26 cm squared. Find the orginal dimensions of the rectan      Log On


   

Question 131750: The lenght of a rectangle is 3 cm greater than the width. If each dimension is increased by 2cm, the area is increased by 26 cm squared. Find the orginal dimensions of the rectangle.
Please send the problem in linear form to Kathy.Christensen@wowway.com
Thank you
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Let x= width
x+3= length
x(x+3) = original area

x+2= increased width
x+3+2=x+5= increased length
(x+2)(x+5)= larger area
x^2+7x+10 = larger area

Original area + 26 = Larger area
x%5E2%2B3x%2B26=x%5E2%2B7x%2B10


Subtract x%5E2 from each side
3x%2B26=7x%2B10
-4x=-16
x=4+cm width
x%2B3=7+cm length

Original rectangle= 4x7=28 sq cm
Increased rectangle = 6x9=54 sq cm

Check: Difference in the area is 26 sq cm.

R^2