The area of a rectangle gets reduced by 9 square units if its length is reducedby 5 units and the breadth is increased by 3 units. If we increasethe length by 3 units and breadth by 2 units, the area is increased by 67square units. Find the length and breadth of the rectangle.

Let the length of the rectangle be x units and the breadth of the rectangle be y units.
Condition 1:
The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and the breadth is increased by 3 units.
(x - 5) (y + 3) = xy - 9
xy + 3x - 5y - 15 = xy - 9
3x - 5y = -9 + 15
3x - 5y = 6… (i)
Condition 2: If we increase the length by 3 units and breadth by 2 units, the area is increased by 67 square units.
(x+3) (y+2) = xy + 67
xy + 2x +3y + 6 = xy + 67
2x + 3y = 67-6
2x + 3y = 61 … (ii)
Multiplying (i) by 3 and (ii) by 5
3 (3x - 5y = 6)
9x - 15y = 18  … (iii)
5 (2x+3y = 61)
10x + 15y = 305 … (iv)
9x - 15y = 18
10x + 15y = 305
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19x           = 323
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x = 323/19
x = 17
Substitute x = 17 in (i)
3x - 5y = 6
3(17) - 5y = 6
51 - 5y = 6
-5y = 6-51
-5y = -45
y = -45/-5
y = 9
. ^. . The length is 17 units and the breadth is 9 units.

Verification:
Area of the rectangle is 17 x 9 = 153 square units
a. (x-5) (y+3) = (17-5) (9+3) = 12 × 12 = 144 = 153 - 9
b. (x+3) (y+2) = (17+3) (9+2) = 20 × 11= 220 = 153 + 67.