### 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 involves two equations with two unknowns, so I might try to solve it by eliminating one of the variables or using substitution.
However, before I do that, I notice that both equations have a similar structure: they are of the form $a+\frac{1}{b}=c.$
This makes me wonder if I can manipulate the equations to create a relationship between $x$ and $y.$
For example, if I multiply both sides of the first equation by $y,$ I get $xy+1=10y.$
Similarly, if I multiply both sides of the second equation by $x,$ I get $xy+1=\frac{5}{12}x.$
Now I have two expressions for $xy+1,$ so I can set them equal to each other and simplify: $10y=\frac{5}{12}x.$
This means that $y=\frac{1}{24}x.$
This is a useful result, because it allows me to replace $y$ with $\frac{1}{24}x$ in either of the original equations and solve for $x.$
I choose the first equation, because it looks simpler: $x+\frac{1}{y}=10.$
Substituting $\frac{1}{24}x$ for $y,$ I get $x+\frac{24}{x}=10.$
This is a quadratic equation, so I can bring all the terms to one side and try to factor or use the quadratic formula: $x^2-10x+24=0.$
Luckily, this equation is easy to factor: $(x-6)(x-4)=0.$
Therefore, $x$ must be either $6$ or $4.$
These are the only possible values for $x,$ because if I plug them back into either of the original equations, I get a consistent value for $y.$
For example, if $x=6,$ then $y=\frac{1}{24}(6)=\frac{1}{4},$ and both equations are satisfied.
Similarly, if $x=4,$ then $y=\frac{1}{24}(4)=\frac{1}{6},$ and both equations are satisfied.