Question 1151633
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x = original length = 3y
y = original breadth or width
A = original area = length*width = x*y = 3y*y = 3y^2


width decreases by 2 ---> y becomes y-2
length increases by ---> x becomes x+4
area is now equal to (new length)*(new width) = (x+4)(y-2)
the old area was 3y^2. One-third of this is y^2 (simply divide by 3), so when it decreases by one-third then we have 3y^2-y^2 = 2y^2 of an area. Basically we have 2/3 of the original area.


(x+4)(y-2) = 2y^2
(3y+4)(y-2) = 2y^2 .... plug in x = 3y
3y(y-2)+4(y-2) = 2y^2 .... distributive rule
3y^2-6y+4y-8 = 2y^2 .... distributive rule
3y^2-2y-8 = 2y^2
3y^2-2y-8-2y^2 = 0
y^2-2y-8 = 0
(y-4)(y+2) = 0 .... factor
y-4 = 0 or y+2 = 0 .... zero product property
y = 4 or y = -2
A negative width does not make sense, so we toss out y = -2 and only focus on y = 4
The original width is 4 meters


Now find the original length
x = 3y
x = 3*4
x = 12


The original length and width of the rectangle is 12 by 4, giving an original area of 12*4 = 48
One-third of this is 48/3 = 16, so if the area decreases by one-third then we have 48-16 = 32 as the new area (note how 32/48 = 2/3). Or you could take 2/3 of 48 and get 32. Either way the new area is 32 square meters.


If we increase the length by 4 and decrease the width by 2, then we have
old length = 12 ----> new length = 16
old width = 4 ---> new width = 2
new area = (new length)*(new width) = 16*2 = 32
which matches up with the previous 32 we got. This helps confirm the answer.
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