Given that,
- $x^{2} – yx – x = 22\quad \longrightarrow(1)$
- $y^{2} – xy + y = 34\quad \longrightarrow(2)$
- $x>y$
From equation $(1),$
$x^{2} – yx – x = 22$
$\Rightarrow x(x-y-1) = 22\quad \longrightarrow(3)$
From equation $(2),$
$y^{2} – xy + y = 34$
$\Rightarrow y(y-x+1) = 34$
Multiply $‘-’$ on both the sides.
$-y(y-x+1) = -34$
$\Rightarrow y(-y+x-1) = -34$
$\Rightarrow y(x-y-1) = -34\quad \longrightarrow(4)$
Subtract equation $(4),$ from equation $(3).$
$ x(x-y-1) - y(x-y-1) = 22-(-34)$
$\Rightarrow (x-y-1)(x-y) = 56 \quad \longrightarrow(5)$
Let, $x-y = m \quad {\color{Blue} {(x>y \Rightarrow x – y >0 \Rightarrow m>0)}}$
Now, $(m-1)m = 56$
$\Rightarrow m^{2}-m-56 =0$
$\Rightarrow m^{2}-8m+7m-56 =0$
$\Rightarrow m(m-8)+7(m-8) =0$
$\Rightarrow (m-8)(m+7) =0$
$\Rightarrow m=8\; \text{(or)}\; m=-7\;{\color{Red} {\text{(rejected)}}}$
$\Rightarrow \boxed{m=8}$
$\Rightarrow \boxed{x-y=8}$
Correct Answer $:\text{B}$