Solve the following system of equations: $\;$ $\dfrac{x + 3}{y - 4} - \dfrac{x - 1}{y + 4} + \dfrac{16}{y^2 - 16} = 0, \;\; 11x - 3y = 1$
Given system of equations:
$\dfrac{x + 3}{y - 4} - \dfrac{x - 1}{y + 4} + \dfrac{16}{y^2 - 16} = 0$ $\;\;\; \cdots \; (1)$
and $\;$ $11x - 3y = 1$ $\;\;\; \cdots \; (2)$
Equation $(1)$ can be written as
$\left(x + 3\right) \left(y + 4\right) - \left(x - 1\right) \left(y - 4\right) + 16 = 0$ $\;\;\;$ (provided $\;$ $y + 4 \neq 0$ $\;$ and $\;$ $y - 4 \neq 0$ $\;$ i.e. $\;$ $y \neq -4$ $\;$ and $\;$ $y \neq 4$)
i.e. $\;$ $xy + 4x + 3y + 12 - xy + 4x + y - 4 + 16 = 0$
i.e. $\;$ $8x + 4y + 24 = 0$
i.e. $\;$ $2x + y + 6 = 0$ $\;\;\; \cdots \; (1a)$
Multiplying equation $(1a)$ with $3$ and equation $(2)$ with $1$ and adding them gives
$17x = - 17$ $\implies$ $x = -1$
Substituting the value of $x$ in equation $(1a)$ gives
$-2 + y + 6 = 0$ $\implies$ $y = -4$
But equation $(1)$ requires $y \neq -4$
$\therefore \;$ The given pair of equations donot have any solution i.e. the solution set is $\phi$ (null set).