**Question:**

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

**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 sum of the two digits of the number is 9 . Thus, we have $x+y=9$

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

Also, 9 times the number is equal to twice the number obtained by reversing the order of the digits. Thus, we have

$9(10 y+x)=2(10 x+y)$

$\Rightarrow 90 y+9 x=20 x+2 y$

$\Rightarrow 20 x+2 y-90 y-9 x=0$

$\Rightarrow 11 x-88 y=0$

$\Rightarrow 11(x-8 y)=0$

$\Rightarrow x-8 y=0$

So, we have the systems of equations

$x+y=9$

$x-8 y=0$

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

Substituting $x=8 y$ from the second equation to the first equation, we get

$8 y+y=9$

$\Rightarrow 9 y=9$

$\Rightarrow y=\frac{9}{9}$

$\Rightarrow y=1$

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

$x-8 \times 1=0$

$\Rightarrow x-8=0$

$\Rightarrow x=8$

Hence, the number is $10 \times 1+8=18$.