Solve for $a,b,c$ :- $a + \frac{1}{b} = \frac{7}{3}$ , $b + \frac{1}{c} = 4$ , $c + \frac{1}{a} = 1$ 
Solve for $a,b,c$ :- $a + \frac{1}{b} = \frac{7}{3}$ , $b + \frac{1}{c} = 4$ , $c + \frac{1}{a} = 1$

What I Tried :- We have $3$ equations , $3$ variables, should be easy .
I did everything, solved these equations but in the end I am getting $3a^2 - 10a + 9 = 0$ . (My work is a bit long so can't show it)
This has complex roots , so this problem just got a little tough .
Can anyone help me?

From the last equation we have $c=1-\frac{1}{a}=\frac{a-1}{a}$ and from the second $b=4-\frac{1}{c}=4-\frac{a}{a-1}==\frac{3a-4}{a-1}$ so the first equation becomes $a+\frac{1}{b}=a+\frac{a-1}{3a-4}=\frac{7}{3}$ or $3a^{2}-4a+a-1=\frac{7}{3}(3a-4)$ so $9a^{2}-9a-3=21a-28$ and hence we have $9a^{2}-30a+25=(3a-5)^{2}=0.$
