Since both expressions equal \(y\), set them equal:

\(x+1=x^2-1\)

\(x^2-x-2=0\)

\((x-2)(x+1)=0\)

So \(x=2\) or \(x=-1\).

Using \(y=x+1\):

If \(x=2\), then \(y=3\).

If \(x=-1\), then \(y=0\).

So the solutions are \((-1,0)\) and \((2,3)\).

Set the two expressions for \(y\) equal:

\(4x+7=2x^2+1\)

\(2x^2-4x-6=0\)

\(x^2-2x-3=0\)

\((x-3)(x+1)=0\)

So \(x=3\) or \(x=-1\).

Using \(y=4x+7\):

If \(x=3\), then \(y=19\).

If \(x=-1\), then \(y=3\).

So the solutions are \((-1,3)\) and \((3,19)\).

Set the expressions for \(y\) equal:

\(x^2-6x+5=x-1\)

\(x^2-7x+6=0\)

\((x-1)(x-6)=0\)

So \(x=1\) or \(x=6\).

Using \(y=x-1\):

If \(x=1\), then \(y=0\).

If \(x=6\), then \(y=5\).

So the solutions are \((1,0)\) and \((6,5)\).

From \(x+y=7\), we get \(y=7-x\).

Set equal:

\(x^2-x-2=7-x\)

\(x^2-9=0\)

\((x-3)(x+3)=0\)

So \(x=3\) or \(x=-3\).

Using \(y=7-x\):

If \(x=3\), then \(y=4\).

If \(x=-3\), then \(y=10\).

So the solutions are \((-3,10)\) and \((3,4)\).

Set the expressions for \(y\) equal:

\(x=6-x^2\)

\(x^2+x-6=0\)

\((x+3)(x-2)=0\)

So \(x=-3\) or \(x=2\).

Using \(y=x\), the corresponding \(y\)-values are the same.

So the solutions are \((-3,-3)\) and \((2,2)\).