Inbrunstlr

2022-10-06

We are required to solve the following system of equations:
$\begin{array}{}\text{(1)}& {x}^{3}+\frac{1}{3{x}^{4}}=5\end{array}$
$\begin{array}{}\text{(2)}& {x}^{4}+\frac{1}{3{x}^{3}}=10\end{array}$
We may multiply (1) by $3{x}^{4}$ throughout and (2) by $3{x}^{3}$ throughout (as 0 is not a solution we may cancel the denominators) to yield.
$\begin{array}{}\text{(3)}& 3{x}^{7}+1=15{x}^{4}\end{array}$
$\begin{array}{}\text{(4)}& 3{x}^{7}+1=30{x}^{3}\end{array}$
Subtracting the two:
$15{x}^{4}-30{x}^{3}=0$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{x}^{3}\left(x-2\right)=$
As $0$ is not a solution we choose $x=2$.
But putting $x=2$ in the original equations does not satisfy them. How come?

odejicahfc

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

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