# Solve the system of equations in the set of real numbers.

Solve the system of equations in the set of real numbers:
$$\begin{cases} \frac1x + \frac1{y+z} = \frac13 \\ \frac1y + \frac1{x+z} = \frac15 \\ \frac1z + \frac1{x+y} = \frac17 \end{cases}$$

I got:

$$\begin{cases} 3(x+y+z)=x(y+z) \\ 5(x+y+z)=y(x+z) \\ 7(x+y+z)=z(x+y) \end{cases}$$

However, no matter how I continue from here, I always get $$x=y=z=0$$, which cannot be true; or I get a new system of equations, but still with 3 variables (which I cannot solve).

How can I solve this problem or how should I approach it?

• To be honest, I haven't bothered to retrace your computations. But why wouldn't concluding that $x=y=z=0$ from the second set of equations be invalid? It might very well be that your first set of equations does not have any solutions... – Card_Trick Dec 18 '18 at 22:40
• @Card_Trick Not necessarily. The multiplication by $x+y+z$ can add extraneous solutions. – Don Thousand Dec 18 '18 at 22:41
• I wasn't questioning that, I was just wondering why Pero was concluding that his conclusion cannot be. – Card_Trick Dec 18 '18 at 22:42
• @Card_Trick Pero is concluding that the $0$ solution is extraneous. – The Great Duck Dec 19 '18 at 5:09

Let $$x+y+z=m$$

Adding all the equations, we get,

$$xy+yz+zx=\frac{3m+5m+7m}{2}=\frac{15m}{2}$$

Subtracting each equation one by one from this, we get,

$$xy=\frac{m}{2}$$

$$yz=\frac{9m}{2}$$

$$zx=\frac{5m}{2}$$

Dividing by $$xyz$$, we get, $$\frac{1}{x}:\frac{1}{y}:\frac{1}{z}=9:5:1$$ $$\Longrightarrow x:y:z=\frac{1}{9}:\frac{1}{5}:\frac{1}{1}$$ $$\Longrightarrow x:y:z=5:9:45$$ Now, let $$x=5k, y=9k, z=45k$$ and get the result.

We know that via your equations, $$\frac{15}2(x+y+z)=xy+yz+xz$$Hence, $$xy=\frac12(x+y+z)$$$$yz=\frac92(x+y+z)$$$$xz=\frac52(x+y+z)$$So, assuming $$x+y+z\neq0$$, $$z=9x$$, $$z=5y$$. Try using this to move forward!
• How do we get $z=9x, z=5y$? – Pero Dec 18 '18 at 22:58