A two-digit number is 4 more than 6 times the sum of its digits. If 18 is subtracted from the number, the digits are reversed. Find the number.
Let the digits at units and tens place of the given number be $x$ and $y$ respectively. Thus, the number is $10 y+x$.
The number is 4 more than 6 times the sum of the two digits. Thus, we have
$\Rightarrow 10 y+x=6 x+6 y+4$
$\Rightarrow 6 x+6 y-10 y-x=-4$
$\Rightarrow 5 x-4 y=-4$
After interchanging the digits, the number becomes $10 x+y$.
If 18 is subtracted from the number, the digits are reversed. Thus, we have
$(10 y+x)-18=10 x+y$
$\Rightarrow 10 x+y-10 y-x=-18$
$\Rightarrow 9 x-9 y=-18$
So, we have the systems of equations
$5 x-4 y=-4$
Hence, the number is $10 \times 6+4=64$.