Solve $x + \frac{ 1 }{y+1/3}=38/3$ in the set of natural numbers

Question

The following equation should have a solution with $x,y$ being natural numbers. I cannot find it. Is there such solution?

$$x + \frac{ 1 }{y+1/3}=\frac{38}{3}$$

Answer 1

Writing the expression as:

$$x+\frac{3}{3y+1}$$ we see that the denominator of the result has to be $3y+1$, and you can't get $3y+1=3$. No solution.

Answer 2

Hint:

$$x+\frac{1}{y+1/3}=38/3$$ $$3x+\frac{9}{3y+1}=38$$ $$3x(3y+1)+9=38(3y+1)$$ $$(3x-38)(3y+1)=-9$$ $$(38-3x)(3y+1)=9$$