Determine the sum of the real numbers $x$ for which $\frac{2 x}{x^{2}+5 x+3}+\frac{3 x}{x^{2}+x+3}=1$.
By clearing denominators, the equation becomes

$$
2 x\left(x^{2}+x+3\right)+3 x\left(x^{2}+5 x+3\right)=\left(x^{2}+5 x+3\right)\left(x^{2}+x+3\right)
$$

and after expanding the left side, we get

$$
2 x^{3}+2 x^{2}+6 x+3 x^{3}+15 x^{2}+9 x=\left(x^{2}+5 x+3\right)\left(x^{2}+x+3\right) \text {. }
$$

There are many ways to systematically expand the right side. One way is to think of it as a binomial

$$
\begin{aligned}
\left(x^{2}+5 x+3\right)\left(x^{2}+x+3\right) & =\left(\left(x^{2}+(5 x+3)\right)\left(x^{2}+(x+3)\right)\right. \\
& =x^{4}+x^{2}(x+3)+(5 x+3) x^{2}+(5 x+3)(x+3) \\
& =x^{4}+x^{3}+3 x^{2}+5 x^{3}+3 x^{2}+5 x^{2}+15 x+3 x+9 \\
& =x^{4}+6 x^{3}+11 x^{2}+18 x+9 .
\end{aligned}
$$

The original equation is equivalent to

$$
2 x^{3}+2 x^{2}+6 x+3 x^{3}+15 x^{2}+9 x=x^{4}+6 x^{3}+11 x^{2}+18 x+9
$$

which can be rearranged to

$$
x^{4}+x^{3}-6 x^{2}+3 x+9=0 \text {. }
$$

Generally, finding the real roots of a quartic polynomial is quite difficult, but the rational roots theorem tells us that any rational roots of the quartic above must be divisors of 9. In general, there could be real roots that are not rational, but that will not be the case here.

The possible rational roots of the quartic are $\pm 1, \pm 3$, and $\pm 9$.

Checking $x=1$, we find that $(1)^{4}+(1)^{3}-6(1)^{2}+3(1)+9=8$, so 1 is not a root.

Checking $x=-1$, we find that $(-1)^{4}+(-1)^{3}-6(-1)^{2}+3(-1)+9=1-1-6-3+9=0$ so -1 is a root.

This means $x+1$ is a factor of the quartic. After factoring, we get that

$$
x^{4}+x^{3}-6 x^{2}+3 x+9=(x+1)\left(x^{3}-6 x+9\right) \text {. }
$$

We now try to find a rational root of $x^{3}-6 x+9$. Again, the possibilities are $\pm 1, \pm 3$, and $\pm 9$. Of these six possibilities, only -3 is a root:

$$
(-3)^{3}-6(-3)+9=-27+18+9=0 .
$$

This means $x+3 \mathrm{~s}$ a factor of $x^{3}-6 x+9$, and after factoring, we get that

$$
x^{4}+x^{3}-6 x^{2}+3 x+9=(x+1)(x+3)\left(x^{2}-3 x+3\right) \text {. }
$$

Thus, the equation we wish to solve is equivalent to

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

The roots of the quartic are therefore $x=-1$ and $x=-3$ and the roots of $x^{2}-3 x+3$.

Using the quadratic formula, the two roots of $x^{2}-3 x+3$ are $\frac{3+\sqrt{-3}}{2}$ and $\frac{3-\sqrt{-3}}{2}$, neither of which are real.

Therefore, the only real values of $x$ that can possibly satisfy the original equation are -3 and -1 .

It can be checked that both $x=-1$ and $x=-3$ satisfy the original equation, so the answer is $(-1)+(-3)=-4$.