Why does this method of solution for this system of equations yield an incorrect answer? We are required to solve the following system of equations:
$$x^3 + \frac{1}{3x^4} = 5 \tag1$$
$$x^4 + \frac{1}{3x^3} = 10 \tag2$$
We may multiply $(1)$ by $3x^4$ throughout and $(2)$ by $3x^3$ throughout (as $0$ is not a solution we may cancel the denominators) to yield. 
$$3x^7 + 1 = 15x^4 \tag3$$
$$3x^7 + 1 = 30x^3 \tag4$$
Subtracting the two:
$$15x^4 - 30x^3 = 0$$
$$\implies x^3(x-2) = $$
As $0$ is not a solution we choose $x=2$.
But putting $x=2$ in the original equations does not satisfy them. How come?

Solutions to (3) and (4) are equivalent to solutions of $15x^4 - 30x^3 = 0$ and one of (3) or (4).  But the solution $x = 2$ to $15 x^4 - 30x^3 = 0$ does not satisfy (3) [or equivalently (4)], as $3*2^7 + 1 \neq 15*2^4$.
You shouldn't be surprised that a system of two equations in one unknown may be inconsistent, i.e. not have any solutions.
