Solve the follwing system of equations for $x, y$ and $z$

Question

$$\frac{y+z}{5}=\frac{z+x}{8}=\frac{x+y}{9}$$ 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.

Answer 1

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

Answer 2

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