X_ 2 (~g)+Y_ 2 (~g) leftharpoons 2 Z(g) X_ 2 (~g) and Y_ 2 (~g) are...

The equilibrium constant for the reaction $X_2(g) + Y_2(g) \rightleftharpoons 2Z(g)$ is calculated from the initial equilibrium state:
$K_c = \frac{[Z]^2}{[X_2][Y_2]} = \frac{9^2}{3 \times 3} = 9$ mol/L.
When 10 mol of Z(g) is added, the new concentrations are: $[X_2] = 3$, $[Y_2] = 3$, $[Z] = 19$ mol.
The reaction quotient is $Q = \frac{19^2}{3 \times 3} = 40.11$.
Since $Q > K_c$, the equilibrium shifts left (toward reactants).
Let $x$ mol of Z decompose. At the new equilibrium: $[X_2] = 3 + \frac{x}{2}$, $[Y_2] = 3 + \frac{x}{2}$, $[Z] = 19 - x$.
Applying the equilibrium expression: $\frac{(19-x)^2}{(3+\frac{x}{2})^2} = 9$.
Taking the square root: $\frac{19-x}{3+\frac{x}{2}} = 3$.
Solving: $19 - x = 9 + \frac{3x}{2}$, which gives $10 = \frac{5x}{2}$, so $x = 4$.
The moles of Z at new equilibrium = $19 - 4 = 15$ mol.