If in a rectangle, the length is increased and breadth reduced each by 2 units,

Solution:

Let the length and breadth of the rectangle be $x$ and $y$ units respectively

Then, area of rectangle $=x y$ square units

If length is increased and breadth reduced each by 2 units, then the area is reduced by 28 square units

$(x+2)(y-2)=x y-28$

$\Rightarrow x y-2 x+2 y-4=x y-28$

$\Rightarrow-2 x+2 y-4+28=0$

$\Rightarrow-2 x+2 y+24=0$

$\Rightarrow 2 x-2 y-24=0$

Therefore, $2 x-2 y-24=0 \cdots(i)$

Then the length is reduced by 1 unit and breadth is increased by 2 units then the area is increased by 33 square units

$(x-1)(y+2)=x y+33$

$\Rightarrow x y+2 x-y-2=x y+33$

$\Rightarrow 2 x-y-2-33=0$

$\Rightarrow 2 x-y-35=0$

Therefore, $2 x-y-35=0 \quad \ldots . .(i i)$

Thus we get the following system of linear equation

$2 x-2 y-24=0$

$2 x-y-35=0$

By using cross multiplication, we have

$\frac{x}{(-2 \times-35)-(-1 \times-24)}=\frac{y}{(2 \times-35)-(2 \times-24)}=\frac{1}{(2 \times-1)-(2 \times-2)}$

$\frac{x}{70-24}=\frac{-y}{-70+48}=\frac{1}{-2+4}$

$\frac{x}{46}=\frac{-y}{-22}=\frac{1}{2}$

$x=\frac{46}{2}$

$x=23$

and

$y=\frac{22}{2}$

$y=11$

The length of rectangle is 23 units.

The breadth of rectangle is 11 units.

Area of rectangle =length $\times$ breadth,

$=x \times y$

$=23 \times 11$

$=253$ square units

Hence, the area of rectangle is 253 square units