[Algebra]
Problem: Compute the sum of all values of $k$ for which there exist positive real numbers $x$ and $y$ satisfying the following system of equations.

$$
\left\{\begin{aligned}
\log _{x} y^{2}+\log _{y} x^{5} & =2 k-1 \\
\log _{x^{2}} y^{5}-\log _{y^{2}} x^{3} & =k-3
\end{aligned}\right.
$$

Solution: Let $\log _{x} y=a$. Then the first equation is equivalent to $2 a+\frac{5}{a}=2 k-1$, and the second equation is equivalent to $\frac{5 a}{2}-\frac{3}{2 a}=k-3$. Solving this system by eliminating $k$ yields the quadratic equation $3 a^{2}+5 a-8=0$, hence $a=1$ or $a=-\frac{8}{3}$. Substituting each of these values



of $a$ into either of the original equations and solving for $k$ yields $(a, k)=(1,4)$ or $\left(-\frac{8}{3},-\frac{149}{48}\right)$. Adding the values of $k$ yields the answer of $43 / 48$.
In terms of $a=\log _{x} y$, the two equations become $2 a+\frac{5}{a}=2 k-1$ and $\frac{5 a}{2}-\frac{3}{2 a}=k-3$. Eliminate $\frac{1}{a}$ to obtain $31 a=16 k-33$; substitute this into either of the original equations and clear denominators to get $96 k^{2}-86 k-1192=0$. The sum of the two roots is $86 / 96=\mathbf{4 3} / \mathbf{4 8}$.

Final answer: $\frac{43}{48}$