If the system of equation $a x + y = 3, x + 2 y = 3, 3 x + 4 y = 7$ is
consistent, then value of $a$ is given by

2

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Solution

The correct option is A $2$
$\begin{matrix} a x + y = 3 \\ (1) \end{matrix}$
$\begin{matrix} x + 2 y = 3 \\ (2) \end{matrix}$
$\begin{matrix} 3 x + 4 y = 7 \\ (3) \end{matrix}$
Solving equation $(2)$ and $(3)$ and substitute the value of $x$ and $y$ in equation
$x = 1, y = 1$
Now $a (1) + 1 = 3 \Rightarrow a = 2$

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