If x + y = 4, x2 + y2 = 14 and x > y. Then the correct value of x and y is?

Answer: Option C

$$\begin{array}{rl}& x^2+y^2=14\\ & x+y=4.............(i)\\ & \text{Squaring both sides}\\ & \Rightarrow x^2+y^2+2xy=16\\ & \Rightarrow 14+2xy=16\\ & \Rightarrow 2xy=2\\ & \Rightarrow xy=1\\ & \Rightarrow x^2+y^2=14\end{array}$$
Subtracting (2xy) from both sides
$$\begin{array}{rl}& \Rightarrow x^2+y^2-2xy=14-2xy\\ & \Rightarrow {\left(x-y\right)}^2=14-2\times 1\\ & \Rightarrow x-y=\sqrt{12}\\ & \Rightarrow x-y=2\sqrt{3}.........(ii)\\ & \text{Solve equation (i) and (ii)}\\ & \Rightarrow y=2-\sqrt{3}\\ & \Rightarrow x=2+\sqrt{3}\end{array}$$