SOLUTION: Someone please help me with this problem! It involves a little geometry as well as algebra I suppose. A rectangle is 5 cm longer than it is wide. If its length and width are b

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Someone please help me with this problem! It involves a little geometry as well as algebra I suppose. A rectangle is 5 cm longer than it is wide. If its length and width are b      Log On


   

Question 126347This question is from textbook Algebra
: Someone please help me with this problem! It involves a little geometry as well as algebra I suppose.
A rectangle is 5 cm longer than it is wide. If its length and width are both increased by 3 cm, its area is increased by 60 cm^2. Find the dimensions of the original rectangle.
I tried to set it up by combining terms to get the equation x^2 + 24 = y + 60 but that produced answers that didn't seem to check.
If you can, please help me out! This question is from textbook Algebra

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let A = original area
Let W = original width
Let L = original length
A+=+W+%2A+L
L+=+W+%2B+5
A+=+W%28W+%2B+5%29
A+=+W%5E2+%2B+5W
A+%2B+60+=+%28W+%2B+3%29%28L+%2B+3%29
A+%2B+60+=+%28W+%2B+3%29%28W+%2B+5+%2B+3%29
A+=+%28W+%2B+3%29%28W+%2B+8%29+-+60
A+=+W%5E2+%2B+11W+%2B+24+-+60
A+=+W%5E2+%2B+11W+-+36
A+=+W%5E2+%2B+5W from above
W%5E2+%2B+5W+=+W%5E2+%2B+11W+-+36
6W+=+36
W+=+6
L+=+W+%2B+5
L+=+11
The dimensions of the original rectangle are 6 x 11 cm2
check
A+=+W+%2A+L
A+=+66
A+%2B+60+=+%28W+%2B+3%29%28L+%2B+3%29
66+%2B+60+=+%286+%2B+3%29%2811+%2B+3%29
126+=+9%2A14
126+=+126
OK