Let a = xy, b=x+y, the n the system of equations is given as

a+b=11

ab=30

According to Vieta's formula, a,b are two roots of the quadratic equation

t^2-11t+30=0

Solving for t gives

\begin{cases} t_1 = 5 \\ t_2=6 \end{cases}

Since a, b are symmetric， permutation of a,b will not change the equality. So we have two cases.

\begin{cases} a = 5 \\ b=6 \end{cases} \quad \text{or} \quad \begin{cases} b = 5 \\ a=6 \end{cases}

Case 1, if a = 5 and b=6, then we have the system

\begin{cases} xy = 5 \\ x+y=6 \end{cases}

Once again, apply the Vieta's formula. x,y is the roots of the quadratic equation

z^2-5z+6=0

Solving the equation yields

\begin{cases} x = 1 \\ y=5 \end{cases} \quad \text{or} \quad \begin{cases} x = 5 \\ y=1 \end{cases}

considering the symmetry of the roots.

Case 2, if a = 65 and b=5, then we have the system

z^2-6z+5=0

Solving the equation yields

\begin{cases} x = 2 \\ y=3 \end{cases} \quad \text{or} \quad \begin{cases} x = 3\\ y=2 \end{cases}

In summary, the solution sets for the system of equations is \{(1,5),(5,1),(2,3),(3,2)\}