<s>[INST] Solve the follwing system of equations for $x, y$ and $z$ $$\frac{y+z}{5}=\frac{z+x}{8}=\frac{x+y}{9}$$ and $$6(x+y+z)=11$$ 
My teacher told me that I would have to get $3$ different equations to get $x, y$ and $z$. I've tried many methods and I'm confused as to how to do this problem.
 [/INST] Hint 
$$\frac{y+z}{5} =\frac{z+x}{8} =  \frac{x+y}{9}$$ is the same as 
$\color{red}{\large{\frac{y+z}{5} =\frac{z+x}{8}}}$ and $\color{red}{\large{\frac{z+x}{8} =  \frac{x+y}{9}}}$
Which is also the same as $\color{blue}{\large{\frac{y+z}{5} = \frac{x+y}{9}}}$ and $\color{blue}{\large{\frac{z+x}{8} = \frac{y+z}{5}}}$
And so you can break it into two equations.
$\color{blue}{\large{\frac{z+x}{8} = \frac{y+z}{5}} \implies 5(z+x)=8(y+z)} \implies 5z + 5x =8y + 8z \implies 5x -8y -3z = 0$
and $\color{blue}{\large{\frac{y+z}{5} = \frac{x+y}{9}}} \implies 9(y+z) = 5(x+y) \implies 9y  + 9z = 5x + 5y \implies 5x + 9z + 4y = 0$
and you have the 3rd equation which is $6(x+y+z) = 11 \implies x + y + z = \frac{11}{6}$.
Now can you solve this system of equations
$$5x -8y -3z = 0, 5x  + 9z + 4y= 0,  x + y + z = \frac{11}{6}$$
Now you can use this link here linear algebra tool kit to check your work
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