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 \((2,2)\).

Set the expressions for \(y\) equal:

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

\(x^2-3x-10=0\)

\((x-5)(x+2)=0\)

So \(x=5\) or \(x=-2\).

Using \(y=x+5\):

If \(x=5\), then \(y=10\).

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

So the solutions are \((-2,3)\) and \((5,10)\).

Set the expressions for \(y\) equal:

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

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

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

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

Using \(y=2x\):

If \(x=1\), then \(y=2\).

If \(x=2\), then \(y=4\).

So the solutions are \((1,2)\) and \((2,4)\).

Set the expressions for \(y\) equal:

\(x^2=4x+5\)

\(x^2-4x-5=0\)

\((x-5)(x+1)=0\)

So \(x=5\) or \(x=-1\).

Using \(y=x^2\):

If \(x=5\), then \(y=25\).

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

So the solutions are \((-1,1)\) and \((5,25)\).

From \(y+3=2x\), we get \(y=2x-3\).

Now set equal:

\(2x-3=x^2-2x+1\)

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

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

So \(x=2\).

Then \(y=2(2)-3=1\).

There is one repeated solution: \((2,1)\).

Set the expressions for \(y\) equal:

\(2x+5=x^2-3x-1\)

\(x^2-5x-6=0\)

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

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

Using \(y=2x+5\):

If \(x=6\), then \(y=17\).

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

So the solutions are \((-1,3)\) and \((6,17)\).

From \(x-2y=8\), we get \(x=8+2y\).

Substitute into \(xy=24\):

\((8+2y)y=24\)

\(2y^2+8y-24=0\)

\(y^2+4y-12=0\)

\((y+6)(y-2)=0\)

So \(y=-6\) or \(y=2\).

If \(y=-6\), then \(x=8+2(-6)=-4\).

If \(y=2\), then \(x=8+4=12\).

So the solutions are \((-4,-6)\) and \((12,2)\).

From \(y+3x=5\), we get \(y=5-3x\).

Substitute into \(x^2+y=3\):

\(x^2+5-3x=3\)

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

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

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

If \(x=1\), then \(y=5-3=2\).

If \(x=2\), then \(y=5-6=-1\).

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

Set the expressions for \(y\) equal:

\(6x+5=12x^2-5x\)

\(12x^2-11x-5=0\)

The factorisation is \((3x+1)(4x-5)=0\).

So \(x=-\dfrac13\) or \(x=\dfrac54\).

Using \(y=6x+5\):

If \(x=-\dfrac13\), then \(y=3\).

If \(x=\dfrac54\), then \(y=\dfrac{25}{2}\).

So the solutions are \(\left(-\dfrac13,3\right)\) and \(\left(\dfrac54,\dfrac{25}{2}\right)\).

From \(2x=y+2\), we get \(y=2x-2\).

Substitute into \(x^2+y^2=8\):

\(x^2+(2x-2)^2=8\)

\(x^2+4x^2-8x+4=8\)

\(5x^2-8x-4=0\)

\((5x+2)(x-2)=0\)

So \(x=-\dfrac25\) or \(x=2\).

If \(x=-\dfrac25\), then \(y=2(-\dfrac25)-2=-\dfrac{14}{5}\).

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

So the solutions are \(\left(-\dfrac25,-\dfrac{14}{5}\right)\) and \((2,2)\).

From \(3x+4y=15\), we get \(y=\dfrac{15-3x}{4}\).

Substitute into \(2xy=9\):

\(2x\cdot \dfrac{15-3x}{4}=9\)

\(x(15-3x)=18\)

\(3x^2-15x+18=0\)

\(x^2-5x+6=0\)

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

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

If \(x=2\), then \(y=\dfrac{9}{4}\).

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

So the solutions are \(\left(2,\dfrac94\right)\) and \(\left(3,\dfrac32\right)\).

From \(3x=1+2y\), we get \(y=\dfrac{3x-1}{2}\).

Substitute into \(3x^2-2y^2+5=0\):

\(3x^2-2\left(\dfrac{3x-1}{2}\right)^2+5=0\).

Multiply through by 2:

\(6x^2-(3x-1)^2+10=0\)

\(6x^2-(9x^2-6x+1)+10=0\)

\(-3x^2+6x+9=0\)

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

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

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

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

If \(x=-1\), then \(y=-2\).

So the solutions are \((-1,-2)\) and \((3,4)\).

Set the expressions for \(y\) equal:

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

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

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

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

Using \(y=x-4\):

If \(x=-3\), then \(y=-7\).

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

So the solutions are \((-3,-7)\) and \((2,-2)\).

Set the expressions for \(y\) equal:

\(x-1=\dfrac12(1-x^2)\)

Multiply by 2:

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

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

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

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

Using \(y=x-1\):

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

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

So the solutions are \((-3,-4)\) and \((1,0)\).

Substitute \(y=1+2x\) into \(y^2=2x^2+x\):

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

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

\(2x^2+3x+1=0\)

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

So \(x=-\dfrac12\) or \(x=-1\).

Using \(y=1+2x\):

If \(x=-\dfrac12\), then \(y=0\).

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

So the solutions are \(\left(-\dfrac12,0\right)\) and \((-1,-1)\).

Notice that

\(x^2-2xy+y^2=(x-y)^2\).

So \((x-y)^2=1\).

Substitute \(y=2x+1\):

\(x-(2x+1)=-x-1\), so

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

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

\(x+1=\pm1\)

So \(x=0\) or \(x=-2\).

Using \(y=2x+1\):

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

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

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

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

\(1+2x-x^2=\dfrac12x+5\)

Multiply through by 2:

\(2+4x-2x^2=x+10\)

\(-2x^2+3x-8=0\)

or \(2x^2-3x+8=0\).

The discriminant is

\(b^2-4ac=(-3)^2-4(2)(8)=9-64=-55\).

Since the discriminant is negative, there are no real solutions.