### Instruction: Given that $x$ and $y$ are nonzero real numbers such that $x+\frac{1}{y}=10$ and $y+\frac{1}{x}=\frac{5}{12},$ find all possible values for $x.$

(Enter your answer as a comma-separated list.).### Response: This problem looks like it involves some algebraic manipulation, so I will start by simplifying the given equations and eliminating fractions.
Multiplying both sides of the first equation by $y,$ I get $xy+1=10y.$
Multiplying both sides of the second equation by $x,$ I get $xy+1=\frac{5x}{12}.$
I notice that the left-hand sides of both equations are the same, so I can set the right-hand sides equal to each other and get $10y=\frac{5x}{12}.$
Solving for $y$ in terms of $x,$ I get $y=\frac{x}{24}.$
Now I can substitute this expression for $y$ into either of the original equations and solve for $x.$
I will use the first equation, since it looks simpler.
Plugging in $y=\frac{x}{24},$ I get $x+\frac{24}{x}=10.$
Multiplying both sides by $x,$ I get $x^2+24=10x.$
Rearranging the terms, I get $x^2-10x+24=0.$
This is a quadratic equation, so I will try to factor it or use the quadratic formula.
I notice that this equation can be factored as $(x-6)(x-4)=0.$
Therefore, the possible values for $x$ are $6$ and $4.$
I will check these values by plugging them back into the original equations and making sure they are consistent.
If $x=6,$ then $y=\frac{6}{24}=\frac{1}{4}.$
Plugging these values into the first equation, I get $6+\frac{1}{\frac{1}{4}}=10,$ which is true.
Plugging these values into the second equation, I get $\frac{1}{4}+\frac{1}{6}=\frac{5}{12},$ which is also true.
So $x=6$ is a valid solution.
If $x=4,$ then $y=\frac{4}{24}=\frac{1}{6}.$
Plugging these values into the first equation, I get $4+\frac{1}{\frac{1}{6}}=10,$ which is true.
Plugging these values into the second equation, I get $\frac{1}{6}+\frac{1}{4}=\frac{5}{12},$ which is also true.
So $x=4$ is a valid solution.