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The sum of a $2$-digit number and the number obtained by reversing the order of its digits, is $121$. The two digits differ by $3$.
(i) Represent the above information in the form of pair of linear equations.
(ii) Show that the equations have unique solution.
(iii) Solve the equations and find the number.
(i) Represent the above information in the form of pair of linear equations.
(ii) Show that the equations have unique solution.
(iii) Solve the equations and find the number.
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Solution: Let the unit digit be $y$ and tens digit be $x (x > y)$
The two-digit number will be $10x + y$
A.T.Q. $(10x + y) + (10y + x) = 121$
(i) $\Rightarrow x + y = 11 \dots (1)$ and $x - y = 3 \dots (2)$
(ii) $\frac{1}{1} \neq \frac{1}{-1}$ therefore equations have unique solution
(iii) Solving equations $(1)$ and $(2)$, we get $x = 7, y = 4 \therefore$ Number is $74$
$47$ may be considered as the correct answer if $y > x$.
The two-digit number will be $10x + y$
A.T.Q. $(10x + y) + (10y + x) = 121$
(i) $\Rightarrow x + y = 11 \dots (1)$ and $x - y = 3 \dots (2)$
(ii) $\frac{1}{1} \neq \frac{1}{-1}$ therefore equations have unique solution
(iii) Solving equations $(1)$ and $(2)$, we get $x = 7, y = 4 \therefore$ Number is $74$
$47$ may be considered as the correct answer if $y > x$.