Key idea. Subtract the two equations to split the system into two cases.

1. Subtraction gives

   \[x^2-y^2+y-x=(x-y)(x+y-1)=0.\]

   Therefore either \(x=y\) or \(x+y=1\).

2. If \(x=y\), then \(x^2+x=12\), so \((x-3)(x+4)=0\). This gives \((3,3)\) and \((-4,-4)\), both of which satisfy the system.

3. If \(x+y=1\), substitute \(y=1-x\):

   \[x^2-x-11=0,\qquad x=\frac{1\pm\sqrt{45}}2=\frac{1\pm3\sqrt5}2.\]

   The value of y uses the opposite sign.

4. The two values in the last case are the roots of \(t^2-t-11=0\) and have sum 1. Hence \(x^2=x+11\) and \(y^2=y+11\), giving \(x^2+y=y^2+x=12\). So these pairs work too.

5. The factored difference forces one of our two cases, so the four pairs found form the complete solution set.

Remember. Subtracting symmetric equations can reveal cases that direct substitution hides.