# SOLUTION: if the length of one side of a square is doubled and the length of an adjacent side is decreased by 3, the area of the resulting rectangle exceeds the area of the original square b

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: if the length of one side of a square is doubled and the length of an adjacent side is decreased by 3, the area of the resulting rectangle exceeds the area of the original square b      Log On

 Ad: Over 600 Algebra Word Problems at edhelper.com Ad: Mathway solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

 Word Problems: Miscellaneous Word Problems Solvers Lessons Answers archive Quiz In Depth

 Question 570236: if the length of one side of a square is doubled and the length of an adjacent side is decreased by 3, the area of the resulting rectangle exceeds the area of the original square by 16. The length of a side of the original square is?Answer by ankor@dixie-net.com(16523)   (Show Source): You can put this solution on YOUR website!if the length of one side of a square is doubled and the length of an adjacent side is decreased by 3, the area of the resulting rectangle exceeds the area of the original square by 16. The length of a side of the original square is? : Let s = length of the side of the original square then s^2 = original area : and the new rectangle dimensions will be 2s by (s-3) : New area - old area = 16 2s(s-3) - s^2 = 16 2s^2 - 6s - s^2 = 16 2s^2 - s^2 - 6s - 16 = 0 s^2 - 6s - 16 = 0 Factors to (s-8)(s+2) = 0 positive solution s = 8 units, the length of the original square : : Check this by finding the areas 16(8-3) = 80 8*8 = 64 ------------- differ; 16