Hello David,
We are given the two equations:
$$x^2+y^2+x+y=530$$
$$xy+x+y=230$$
If we solve the second equation for $x$, we obtain:
$$x=\frac{230-y}{y+1}$$
And then substituting for $x$ into the first equation, we get:
$$\left(\frac{230-y}{y+1}\right)^2+y^2+\frac{230-y}{y+1}+y=530$$
Multiplying through by $(y+1)^2$, and factoring a little and rearranging, there results:
$$(230-y)^2+y(y+1)^3+(230-y)(y+1)-530(y+1)^2=0$$
Distributing and collecting like terms, we obtain:
$$y^4+3y^3-527y^2-1290y+52600=0$$
Utilizing the rational roots theorem, we find that $y=10$ and $y=20$ are roots, and performing the division, we find the equation may be factored as:
$$(y-20)(y-10)\left(y^2+33y+263\right)=0$$
Using the quadratic formula on the quadratic factor, we find the remaining two roots:
$$y=\frac{-33\pm\sqrt{37}}{2}$$
And thus, using the formula for $x$ as a function of $y$ we found earlier, we find the four solutions:
$$\bbox[10px,border:2px solid #207498]{(x,y)=(10,20),\,(20,10),\,\left(-\frac{33+\sqrt{37}}{2},\frac{-33+\sqrt{37}}{2}\right),\,\left(\frac{-33+\sqrt{37}}{2},-\frac{33+\sqrt{37}}{2}\right)}$$