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(C) For a pair of linear equations to have a unique solution $(x=2, y=-3)$,these values must satisfy both equations in the pair.Let us check the optio

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$x-4y-14=0$ and $5x-y-13=0$

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(C) For a pair of linear equations to have a unique solution $(x=2, y=-3)$,these values must satisfy both equations in the pair.Let us check the options:Option $(A): x+y=1 \implies 2+(-3) = -1 
eq 1$. (Incorrect)Option $(B): 2x-y=1 \implies 2(2)-(-3) = 4+3 = 7 
eq 1$. (Incorrect)Option $(C): x-4y-14=0 \implies 2-4(-3)-14 = 2+12-14 = 0$. (Satisfied)$5x-y-13=0 \implies 5(2)-(-3)-13 = 10+3-13 = 0$. (Satisfied)Since both equations in option $(C)$ are satisfied by $(x=2, y=-3)$ and the lines are not coincident (slopes are $1/4$ and $5$),the solution is unique.Note: Option $(D)$ represents coincident lines ($4x+10y=-22$ is $2$ times $2x+5y=-11$),which have infinitely many solutions,not a unique one.

$A$ pair of linear equations which has a unique solution $x=2, y=-3$ is

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