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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 $:\text{B}$

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