most common practice is when we type simultaneous equations in form like that: \begin{cases} x + y = 2 \\ x - y = 2 \\ \end{cases} Related question - What is the notation for a system of simultaneous equations?

But, what if have equation like $$ x^4 - 5x^2 + 6 =0 $$ and i want to denote it into 2 equations: $$x^2=2$$ and $$x^2=3$$ They both will be the solutions but not simultaneous, I mean if I say that $x=+\sqrt3$ it will be one of the solutions for 2nd and original equations, but not for 1st.

I saw notations with square brackets, like is that TeX question or with using word $ or $ between them.

So how can I correctly write such equations?

  • $\begingroup$ $(x^2-2)(x^2-3)=0$ $\endgroup$ – David G. Stork Apr 24 at 18:06
  • $\begingroup$ @DavidG.Stork, that looks like just factoring equation. What if I'll have more complex equations in such factors, which I'll need to process with additional computation? $\endgroup$ – Zekfad Apr 24 at 18:09
  • $\begingroup$ For more complicated equations in more variables Algebraic Geometry has concepts which capture the kind of thing you are describing (Zariski topology). In single variable generalisations the concepts of Galois theory might be relevant. In each case the roots are mapped onto parallel structures whose decomposition reflects the structure inherent in the original equation. $\endgroup$ – Mark Bennet Apr 24 at 18:15

"$x^2=2$ or $x^2=3$" is perfectly acceptable notation.

| cite | improve this answer | |

Also in the first example, we can write both equations also as one equation $(x+y-2)(x-y-2)=0$. However, the system of two equations need not be equivalent to the one given equation, i.e., $x^4-5x^2+6=0$ is not equivalent to $x^2-2=0$ and $x^2-3=0$, since $(x^2-2)(x^2-3)=0$ only implies that one factor must be zero, but not both.

In general, we have here systems of (polynomial) equations. For example,

\begin{align*} xy^2z + 27xz^4 - y^5 & =0\\ x^3yz -xyz +20 & =0\\ xy^3z+xz^2+x+y+z & =0 \end{align*}

A solution has to satisfy all equations simultaneously. Your systems are \begin{align*} x+y-2 & =0,\\ x-y-2 & = 0, \end{align*} and \begin{align*} (x^2-2)(x^2-3) & =0, \end{align*}

| cite | improve this answer | |
  • $\begingroup$ That's pretty helpful, thanks! But still, how can I process after that? E.g. $(x^2-2)=0$ (missing piece) $(x^2-3)=0$ and split my paper into 2 parts? $\endgroup$ – Zekfad Apr 24 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.