<s>[INST] Solve the equation: $\frac{z}{z-5}+\frac{1}{3}=-\frac{5}{5-z}$ Solve the equation: $\frac{z}{z-5}+\frac{1}{3}=-\frac{5}{5-z}$
First $z$ cannot be equal to $5$.
First, I multiplied $z$ with $3$, $1$ with $z-5$ and $-5$ with both. Eliminating the denominators gives me:$3z + z-5 = - 15$. Simplified, I'm left with: $4z-5=-15$. 
Somehow I don't see how this can be solved. If it were $15$, I could at least say that $4\cdot5-5=15$. But that there aren't any solutions because $z$ cannot be $5$. 
What am I not seeing?
 [/INST] This equation has no solution. Change sides to see that 
$$\frac{z}{z-5}+(1/3)+\frac{5}{5-z}=0 \\
Or,\frac{z}{z-5}+(1/3)+\frac{-5}{z-5}=0 \\
Or, \frac{z-5}{z-5}+(1/3)=0$$
Which is impossible.
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