A two-digit number is 4 more than 6 times the sum of its digits.

Solution:

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

$10 y+x=6(x+y)+4$

$\Rightarrow 10 y+x=6 x+6 y+4$

$\Rightarrow 6 x+6 y-10 y-x=-4$

 

$\Rightarrow 5 x-4 y=-4$

After interchanging the digits, the number becomes $10 x+y$.

If 18 is subtracted from the number, the digits are reversed. Thus, we have

$(10 y+x)-18=10 x+y$

$\Rightarrow 10 x+y-10 y-x=-18$

$\Rightarrow 9 x-9 y=-18$

 

$\Rightarrow 9(x-y)=-18$

$\Rightarrow x-y=-\frac{18}{9}$

$\Rightarrow x-y=-2$

So, we have the systems of equations

$5 x-4 y=-4$

 

$x-y=-2$

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Multiplying the second equation by 5 and then subtracting from the first, we have

$(5 x-4 y)-(5 x-5 y)=-4-(-2 \times 5)$

$\Rightarrow 5 x-4 y-5 x+5 y=-4+10$

$\Rightarrow y=6$

Substituting the value of y in the second equation, we have

$x-6=-2$

$\Rightarrow x=6-2$

$\Rightarrow x=4$

Hence, the number is $10 \times 6+4=64$.