Q2(iii):

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (iii) 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.

Given:

1. A two-digit number where the sum of its digits is $9$.

2. Nine times the original number is equal to twice the number obtained by reversing the digits.

To Find:

The original two-digit number.

Step 1: Defining the Variables

Let the digit at the tens place be $x$ and the digit at the units place be $y$.

Since it is a two-digit number, the value of the number can be expressed as:

Original Number $= 10x + y$

When the digits are reversed, the new tens digit becomes $y$ and the new units digit becomes $x$.

Reversed Number $= 10y + x$

Step 2: Formulating the Equations

According to the first condition, the sum of the digits is $9$:

$x + y = 9$ --- (Equation 1)

According to the second condition, nine times the original number is twice the reversed number:

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

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

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

$88x - 11y = 0$

Dividing the entire equation by $11$ to simplify:

$8x - y = 0$ --- (Equation 2)

Step 3: Solving by the Elimination Method

We have the system of equations:

(1) $x + y = 9$

(2) $8x - y = 0$

To eliminate $y$, we add Equation 1 and Equation 2:

$(x + y) + (8x - y) = 9 + 0$

$x + 8x + y - y = 9$

$9x = 9$

$x = \frac{9}{9}$

$x = 1$

Step 4: Finding the value of $y$

Substitute $x = 1$ into Equation 1:

$1 + y = 9$

$y = 9 - 1$

$y = 8$

Step 5: Determining the Number

The tens digit $x = 1$ and the units digit $y = 8$.

Original Number $= 10x + y = 10(1) + 8 = 18$.

Verification:

Sum of digits: $1 + 8 = 9$ (Satisfied).

Nine times the number: $9 \times 18 = 162$.

Twice the reversed number: $2 \times 81 = 162$.

Since $162 = 162$, the solution is correct.

Final Answer: The two-digit number is 18.