cocassi function algebra 2

Let x+y=4xy ..........(1) y+z=6yz ...........(2) z+x=8zx .............(3) Divide (1) by xy, Divide (2) by yz, Divide (3) by zx. Then you will get three equations

1.(1/x)+(1/y)=4

2.(1/y)+(1/z)=6

3.(1/z)+(1/y)=8

Three equation three variable and you will get x=1/3,y=1,z=1/5 x+y+z=23/15

Hence the answer is 23+15=38

I solved for $y$ in terms of $x$ in the first equation: $y=\frac{x}{4x-1}.$ Then I solved for $z$ in terms of $y$ in the second equation: $z=\frac{y}{6y-1}.$ Substituting the first equation into the second gives the complex fraction: $z=\frac{\frac{x}{4x-1}}{\frac{6x}{4x-1}-1}.$ After simplifying, I got $z=\frac{x}{2x+1}.$ Next, I substituted $\frac{x}{2x+1}$ in for $z$ in the third equation: $\frac{x}{2x+1}+x=8\frac{x}{2x+1}x$ $\Rightarrow x+2x^2+x=8x^2$ $\Rightarrow 2x = 6x^2$ $\Rightarrow x=\frac{1}{3}.$ Finally, substituting $x=\frac{1}{3}$ in $y=\frac{x}{4x-1}$ and $z=\frac{x}{2x+1}$ , I get the final answer

$(x,y,z)=(\frac{1}{3},1,\frac{1}{5})$ .