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.
 A: assume that 
$$\frac{y+z}{5} = \frac{z+x}{8} = \frac{x+y}{9} = k$$ 
    where k is a real number
now you can write three equations by cross-multiplying each denominator by k, that yields.
(1) y + z = 5k
(2) z + x = 8k
(3) x + y = 9k
when you add all of these three equations you will get
2(x + y + z) = 22k
i.e. (a) x + y + z = 11k
but you already have an equation that says 
 6(x + y + z) = 11
i.e. (b) x + y + z = 11/6
there fore by considering (a) and (b) you can say that k = 1/6
now we have to find the values of x, y and z.
(1) - (2) implies
 (4) y - x = -3k
(4) + (3) implies
2y = 6k 
i.e y = 3k = 3 x 1/6 = 1/2
there fore y = 1/2
(3) implies 
x + y = 9k = 9 x 1/6 = 3/2
but y = 1/2
so x = 1
(2) implies
x + z = 8k = 8 x 1/6 = 4/3
but x = 1
so z = 1/3
A: 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
