The length of a rectangle exceeds its breadth by 7 cm. If the length is decreased by 4 cm and the breadth is increased by 3 cm, the area of the new rectangle is the same as the area of the original rectangle. Find the length and the breadth of the original rectangle.
Let the breadth of the original rectangle be $\mathrm{x} \mathrm{cm}$.
Then, its length will be $(\mathrm{x}+7) \mathrm{cm}$.
The area of the rectangle will be $(\mathrm{x})(\mathrm{x}+7) \mathrm{cm}^{2}$.
$\therefore(x+3)(x+7-4)=(x)(x+7)$
$\Rightarrow(x+3)(x+3)=x^{2}+7 x$
$\Rightarrow x^{2}+3 x+3 x+9=x^{2}+7 x$
$\Rightarrow x^{2}+6 x+9=x^{2}+7 x$
$\Rightarrow 9=x^{2}-x^{2}+7 x-6 x$
$\Rightarrow 9=x$
$\Rightarrow x=9 \quad$ (by transposition)
Breadth of the original rectangle $=9 \mathrm{~cm}$
Length of the original rectangle $=(\mathrm{x}+7)=(9+7)=16 \mathrm{~cm}$