If $$-\left(\frac{3}{4}\right)x + 3 y - \left(\frac{1}{2}\right) = \left(\frac{3}{2}\right) y -\left(\frac{1}{4}\right) x$$
What is the value of $$x$$ ?
I. $$y^2 = 4$$
II. $$y = 2$$

Expression : $$-\left(\frac{3}{4}\right)x + 3 y - \left(\frac{1}{2}\right) = \left(\frac{3}{2}\right) y -\left(\frac{1}{4}\right) x$$ -------------(i)

I : $$y^2=4$$

=> $$y=\pm2$$

Case 1 : $$y=2$$ substituting in equation (i)

=> $$-(\frac{3}{4})x + 3 (2) - (\frac{1}{2}) = (\frac{3}{2}) \times(2) -(\frac{1}{4}) x$$

=> $$\frac{-2x}{4}=3-6+\frac{1}{2}$$

=> $$\frac{-x}{2}=\frac{-5}{2}$$

=> $$x=5$$

Case 2 : $$y=-2$$ substituting in equation (i)

=> $$-(\frac{3}{4})x + 3 (-2) - (\frac{1}{2}) = (\frac{3}{2}) \times(-2) -(\frac{1}{4}) x$$

=> $$\frac{-2x}{4}=-3+6+\frac{1}{2}$$

=> $$\frac{-x}{2}=\frac{7}{2}$$

=> $$x=-7$$

$$\because$$ Both cases have different values for $$x$$, thus, we cannot find value of $$x$$ using statement I alone.

II : $$y=2$$ substituting in equation (i)

=> $$-(\frac{3}{4})x + 3 (2) - (\frac{1}{2}) = (\frac{3}{2}) \times(2) -(\frac{1}{4}) x$$

=> $$\frac{-2x}{4}=3-6+\frac{1}{2}$$

=> $$\frac{-x}{2}=\frac{-5}{2}$$

=> $$x=5$$

$$\therefore$$ Statement II alone is sufficient.

=> Ans - (B)