Question

Form the pair of linear equations in the following problem and find their solution (if they exist) by any algebraic method:

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.

Answer 1

Step 1: Form the linear equations

Let us assume that the unit digit and tens digit of a number be $x$ and $y$ respectively.

Then, Number $\left(n\right)=10B+A$.

$n$ after reversing the order of the digits $=10A+B$.

According to the given information,

$A+B=9\dots \dots \dots \dots \dots \dots \dots \dots .\left(i\right)$

$9(10B+A)=2(10A+B)$

$\Rightarrow$ $90B+9A=20A+2B$

$\Rightarrow$$88B–11A=0$

$\Rightarrow$ $-A+8B=0\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(ii\right)$

Step 2: Find the solution

Adding the equations $\left(i\right)$ and $\left(ii\right)$ we get,

$A+B-A+8B=9$

$\Rightarrow$ $9B=9$

$\Rightarrow$ $B=1$

Substituting this value of $B$, in the equation $\left(i\right)$ we get

$A+1=9$

$\Rightarrow$$A=8$

$\therefore$The number $\left(n\right)=10B+A$

$=10\times 1+8$

$=18$

Hence, the number $\left(n\right)$ is $18$.