This is going to seem like a highly obscure question because I've likely missed something fairly obvious. Apologies in advance.

Consider the following equations

$$2k+2mk=k$$ $$k+3mk=mk$$

which must be valid for all real values of $k$.

Now obviously the two equations are equivalent and common sense would say that there is only one value of m that satisfies these equations, namely $m=-\frac{1}{2}$, which can be obtained by combining the two equations using addition or simply solving one equation.

However, if I divide equation 1 by equation 2, or vice versa, I end up with a quadratic which has solutions $m=-\frac{1}{2}, m=1$. Clearly, the latter is incorrect.

Could someone please clarify why these equations (or any equations two equations of the same sort) cannot be divided? Addition is possible because you're doing the same thing to both sides of whichever equation when you combine them. Should division not also work, as the ratios obtained would be equal?

The topic area is linear transformations, but I'm looking for a more general answer.

  • $\begingroup$ The equations are not identical, though they might be equivalent. Either way, you’re potentially dividing by zero. $\endgroup$ – amd May 7 '18 at 18:32
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    $\begingroup$ Think of $x = 1$ After squaring both sides we have $x^2= 1$ which has two solutions $(\pm 1)$ The fact is that squaring or submitting to nonlinear operations we can preserve the equality but we can introduce also extraneous solutions... $\endgroup$ – Cesareo May 7 '18 at 18:33
  • $\begingroup$ Thanks, I've made the correction. $\endgroup$ – s.xw May 7 '18 at 18:35
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    $\begingroup$ @s.xw Regarding the division by zero point: in fact, in this example, dividing by zero is eliminating solutions, not adding extra ones! Considering the system of equations in the variables $(m,k)$, in addition to $(m,k) = (-\frac12,k)$, which is a valid solution for any $k$, the solution $(m,0)$ is valid for any $m$. By dividing the two equations you eliminate all the solutions $(m,0)$ except the $(1,0)$ solutions. $\endgroup$ – BallBoy May 7 '18 at 18:56
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    $\begingroup$ Note that after the division you follow with a multiplication... $\endgroup$ – Cesareo May 7 '18 at 18:56

Now obviously the two equations are equivalent

They are indeed equivalent, so you can ignore the second equation, and focus on just the first one:

$$ 2k+2mk=k \quad\iff\quad 2mk+2k-k=0 \quad\iff\quad (2m+1)k=0 $$

The latter can only hold true for all values of $\,k\,$ iff the LHS is the zero polynomial i.e.$\,2m+1=0\,$.

Could someone please clarify why these equations (or any equations two equations of the same sort) cannot be divided?

Because, in general, "conflating" several equations into one is prone to introducing extraneous solutions. Simple example, similar to OP's case, where of course the only actual solution is $\,x=1\,$:

$$ \begin{cases} \begin{align} x &= 1 \\ 2x &= x+1 \end{align} \end{cases} \;\;\implies\;\; x \cdot 2 x = 1 \cdot (x+1) \;\;\implies\;\; 2x^2 - x - 1 = 0 \;\;\implies\;\; x \in \{\color{red}{-\frac{1}{2}}, 1\} $$


$$6k+6mk = 3k $$ $$ 2k + 6mk = 2mk $$ subtract $$ 4k = (3-2m)k $$ $$ (2m+1) k = 0. $$ when $k$ is not zero, we need $m = -1/2$

Better not to divide anything here


If you multiply the equations after dividing both sides by k: $(2 + 2m)(1 + 3m) = m$

$6m^2 + 7m + 2 = 0$

$m = \frac{-1}{2}$ and $m = \frac{-2}{3}$

which both work for $(2 + 2m)(1 + 3m) = m$

However, factoring $(2 + 2m)(1 + 3m) = m$ back to separate equations can result in more than one outcome

$(2 + 2m) = 1$ and $(1 + 3m) = m$ where $m = \frac{-1}{2}$ works for the original equations

and $(2 + 2m) = -m$ and $(1 + 3m) = -1$ where $m = \frac{-2}{3}$ works for a different set.


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