Question:
A two-digit number is 4 times the sum of its digits and twice the product of its digit. Find the number.
Solution:
Let the digits at units and tens places be $x$ and $y$, respectively.
$\therefore$ Original number $=10 y+x$
According to the question:
$10 y+x=4(x+y)$
$\Rightarrow 10 y+x=4 x+4 y$
$\Rightarrow 3 x-6 y=0$
$\Rightarrow 3 x=6 y$
$\Rightarrow x=2 y \ldots .(\mathrm{i})$
Also,
$10 y+x=2 x y$
$\Rightarrow 10 y+2 y=2.2 y . y \quad[$ From (i) $]$
$\Rightarrow 12 y=4 y^{2}$
$\Rightarrow y=3$
From (i), we get:
$x=2 \times 3=6$
$\therefore$ Original number $=10 \times 3+6=36$