SOLUTION: A rectangle is 5 cm. longer than it is wide. If its length and width both increased by 3 cm. , its area is increased by 60 cm (squared). Find the dimensions of the original rectang

Algebra ->  Expressions -> SOLUTION: A rectangle is 5 cm. longer than it is wide. If its length and width both increased by 3 cm. , its area is increased by 60 cm (squared). Find the dimensions of the original rectang      Log On


   

Question 166767: A rectangle is 5 cm. longer than it is wide. If its length and width both increased by 3 cm. , its area is increased by 60 cm (squared). Find the dimensions of the original rectangle.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let y= length in cm
Let x= width in cm
Let A= area
A+=+xy
It is given that
y+=+x+%2B+5
A+=+x%2A%28x%2B5%29
(1) A+=+x%5E2+%2B+5x
Also given is
A+%2B+60+=+%28x+%2B+3%29%28y+%2B+3%29
And, since y+=+x+%2B+5
A+%2B+60+=+%28x+%2B+3%29%28x+%2B+8%29
A+%2B+60+=+x%5E2+%2B+11x+%2B+24
(2) A+=+x%5E2+%2B+11x+-+36
Since (1) and (2) are both equations
for A, make them equal to eachother
x%5E2+%2B+5x+=+x%5E2+%2B+11x+-+36
5x+=+11x+-+36
6x+=+36
x+=+6
y+=+x+%2B+5
y+=+6+%2B+5
y+=+11
The dimensions of the original rectangle are 11 x 6 cm
check:
A+=+xy
A+=+66 cm2
A+%2B+60+=+%28x+%2B+3%29%28y+%2B+3%29
A+%2B+60+=+9%2A14
A+=+126+-+60
A+=+66 cm2
OK