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.

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
