SOLUTION: The problem is: A rectangle is 5 cm longer than it is wide. If its length and width are both incrased by 3 cm, its area is increased by 60 cm^2. Find the dimensions of the ori

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: The problem is: A rectangle is 5 cm longer than it is wide. If its length and width are both incrased by 3 cm, its area is increased by 60 cm^2. Find the dimensions of the ori      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 129641This question is from textbook Algebra Structure and Method
:
The problem is: A rectangle is 5 cm longer than it is wide. If its length and width are both incrased by 3 cm, its area is increased by 60 cm^2. Find the dimensions of the original rectangle
I have completely confused myself. Here is what I have done:
A[1] = x(x+5)
A[2} = (x+3)(x+5)
A[1] = (x+3)(x+5)-60
x^2 + 5x = (x+3)(x+5)-60
x^2 +5x = x^2 +5x + 3x +15 - 60
5x = 8x - 45
-5x = -8x + 45
3x = 45
x = 15
x + 5 = 20
Dimensions of original rectangle are 15, 20.
Answer key says correct answer is x = 6, x+5 = 11.
This question is from textbook Algebra Structure and Method

Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
ORIGINAL RECTANGLE=X(X+5)=X^2+5X
EXPANDED RECTANGLE=(X+3)(X+8)=X^2+11X+24 HERE YOU HAVE TO INCREASE BOTH THE LENGTH & THE WIDTH.
X^2+5X=X^2+11X+24-60
5X=11X-36
5X-11X=-36
-6X=-36
X=-36/-6
X=6 THE ORIGINAL WIDTH.
6+5=11 THE ORIGINAL LENGTH.
PROOF:
6*11=(6+3)(6+8)-60
66=9*14-60
66=126-60
66=66