Solve the following system of equations and investigate them with respect to $a$: $\;$ $2x + 3y = 5$, $\;$ $x - y = 2$, $\;$ $x + 4y = a$
Given system of equations:
$2x + 3y = 5$ $\;\;\; \cdots \; (1)$, $\;\;$ $x - y = 2$ $\;\;\; \cdots \; (2)$, $\; \;$ $x + 4y = a$ $\;\;\; \cdots \; (3)$
Multiplying equation $(2)$ with $3$ gives
$3x - 3y = 6$ $\;\;\; \cdots \; (4)$
Adding equations $(1)$ and $(4)$ gives
$5x = 11$ $\implies$ $x = \dfrac{11}{5}$
Substituting the value of $x$ in equation $(2)$ gives
$y = \dfrac{11}{5} - 2 = \dfrac{1}{5}$
Substituting the values of $x$ and $y$ in equation $(3)$ gives
$\dfrac{11}{5} + \dfrac{4}{5} = a$ $\implies$ $a = \dfrac{15}{5} = 3$
$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(\dfrac{11}{5}, \dfrac{1}{5}\right) \right\}$ $\;$ when $\;$ $a = 3$
When $\;$ $a \neq 3$, $\;$ then equation $(3)$ is not satisfied and therefore the solution of the given system of equations is $\phi$.