Question 228677: The length of a rectangle is three less than twice its width. If the length is increased by 3 and the width is decreased by 1, a new rectangle is formed whose area is 5 more than the original rectangle. Find:
a) The dimensions of the original rectangle.
b) The area of the new rectangle.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! The length of a rectangle is three less than twice its width. If the length is increased by 3 and the width is decreased by 1, a new rectangle is formed whose area is 5 more than the original rectangle.
Find:
a) The dimensions of the original rectangle.
Let x = the width
then
(2x-3) = the length
and
A = x(2x-3)
A = 2x^2 - 3x; area of original
;
b) The area of the new rectangle.
(x-1) = the width of the new rectangle
and
(2x-3) + 3 = 2x; length of the new rectangle
and
A = 2x(x-1)
A = 2x^2 - 2x; area of new
:
"new rectangle is formed whose area is 5 more than the original rectangle.
New area - old area = 5
2x^2 - 2x - (2x^2 - 3x) = 5
2x^2 - 2x - 2x^2 + 3x = 5
2x^2 - 2x^2 - 2x + 3x = 5
x = 5 units, original width
then
2(5) - 3 = 7 units, original length
and
5 - 1 = 4 units, new width
then
2(5) = 10 units, new length
;
;
Check solution using the area statement
new area - original area
10*4 - 7*5 = 5
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