### 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: I notice that both equations have the same structure: they involve a variable plus the reciprocal of another variable.
I wonder if I can use this similarity to eliminate one of the variables and solve for the other.
I try multiplying the first equation by $y$ and the second equation by $x.$
This gives me $xy+1=10y$ and $yx+1=\frac{5}{12}x.$
I see that both equations have $xy+1$ in them, so I can set them equal to each other and get $10y=\frac{5}{12}x.$
I solve for $y$ in terms of $x$ by dividing both sides by $10.$
This gives me $y=\frac{1}{24}x.$
I plug this expression for $y$ into the first equation and get $x+\frac{24}{x}=10.$
I want to get rid of the fraction, so I multiply both sides by $x.$
This gives me $x^2+24=10x.$
I rearrange this into a quadratic equation by subtracting $10x$ from both sides.
This gives me $x^2-10x+24=0.$
I factor this equation by looking for two numbers that multiply to $24$ and add to $-10.$
I find that $-6$ and $-4$ work, so I write $x^2-10x+24=(x-6)(x-4)=0.$
I use the zero product property and conclude that $x-6=0$ or $x-4=0.$
Therefore, $x=6$ or $x=4.$
These are the two possible values for $x.$