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

Solution(By Examveda Team)

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