0

I was trying to solve this system of equations:

$${ \begin{cases}{x^4+y^4=1} \\ {x^3+y^3=1}\end{cases} }$$ I subtracted the second equation from the first one:$$x^4-x^3+y^4-y^3=0$$ $$x^3(x-1)+y^3(y-1)=0$$ We can easily check that $(x,y)=(1,1)$ and $(x,y)=(0,0)$ are roots of this equation. but if we plug in these values in the original system of equations we can see they are not the answer. So why this is happening?

Etemon
  • 6,437
  • 2
    Say we have $a= 1$ and $b=1$. Subtracting the second equation from the first we have $a-b=0$, which is satisfied by $(0,0)$, but is not the solution of the original system. – player3236 Feb 10 '21 at 15:06
  • 1
    @player3236 But I saw a lot of system of equations that we subtract equations and continue from that so this is known method in solving system of equations. so is there anything we should consider to avoid this error when we are subtracting? – Etemon Feb 10 '21 at 15:09
  • 2
    Now that is a good question. The moral is that even if we have the new equation, we must put the original equation(s) into consideration. – player3236 Feb 10 '21 at 15:13
  • 1
    Extraneous solutions: wikipedia, MSE – ryang Feb 10 '21 at 15:16
  • 2
    @Soheil $\begin{cases}A=B\ C=D\end{cases}$ is equivalent to $\begin{cases}A-C=B-D\ C=D\end{cases}$, but it isn't equivalent to $A-C=B-D$. I don't quite know where you could have found otherwise, because in 90% of the sensible cases you just lose a condition when you only consider $A-C=B-D$. –  Feb 10 '21 at 15:16
  • @Gae.S. Oh I see. thank you very much. – Etemon Feb 10 '21 at 15:18
  • 2
    In my opinion, the direction of the implication(s) in attacking problems like this (or similar problems where you square both sides) is often misunderstood. In general, as your analysis goes from eqn (1) to eqn (2) to eqn (3), the valid inference is that if $S$ is a solution to eqn (n), then it is a solution to eqn (n+1). This means that the inferences are often one-way implications, rather than the occasional two-way implications. That means that the derived solutions are typically nothing more than candidate solutions that must be checked against the original equation(s). – user2661923 Feb 10 '21 at 15:21
  • 1
    As an extra remark, notice that $(0,0)$ and $(1,1)$ aren't the only solutions to the equation $x^3(x-1)+y^3(y-1)=0$. Specifically, for all $\alpha\in \left[-\frac{81}{256},0\right)$ we have that $#{u\in\Bbb R,:, u^3(u-1)=\alpha}\ge1$ and $#{u\in\Bbb R,:, u^3(u-1)=-\alpha}=2$, so you can take appropriate $x$ and $y$ from those sets. –  Feb 10 '21 at 15:30

1 Answers1

2

Your new equation is indeed valid, but it alone doesn't determine solutions to the original equations. You must still have a second companion equation.

For example, look at the first original equation. There are lots of solutions to that equation, but they aren't necessarily solutions to the second original equation as well.

Any solution of the original system must also satisfy your new equation. But only those solutions which satisfy both your new equation and a companion equation (whatever you use for the new second equation) will also be solutions of the original equations, provided the process of obtaining the two new equations is reversible (such as making linear combinations of the two old equations, which you did).

MPW
  • 43,638