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I want to know what can happen if we multiply both sides of an equation by $\frac{1}{x}$, where $x$ is a variable.

I mean, is it possible that we get redundant equations? Or defective equations ?

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    $\begingroup$ Please give more specific information. Th only precaution I can give you here is that when multiplying both sides by 1/x, you have to give the condiftion that x is not equal to zero. $\endgroup$ – Shikhar S. Maheshwari Apr 30 '16 at 9:40
  • $\begingroup$ Your question is not clear. How can you have "redundant equations" when there is only one base equation? In what way can an equation be "defective"? $\endgroup$ – Rory Daulton Apr 30 '16 at 10:17
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    $\begingroup$ I mean given an equation, is it possible that we get a redundant equation by multiplying both sides by 1/x? The definition of defective equation is an equation that has fewer solutions than the original. $\endgroup$ – mick Apr 30 '16 at 22:07
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You might find it useful to argue "by cases". An example would be instructive:

Find all $x$ satisfying $x^2 + x = 0$.

Seeing the common factor of $x$, it is tempting to divide both sides by $x$ (i.e., multiply both sides by $\frac{1}{x}$). Any time you divide by an unknown quantity, however, you should be concerned about the case where that quantity is zero (since division by zero is undefined). You could use cases to organize your thinking: either $x$ is zero or it is not.

Case 1: $x = 0$

Since $x = 0$, we can substitue to find $0^2 + 0$ is indeed $0$, so $x = 0$ is a solution.

Case 2: $x \neq 0$

Since $x \neq 0$, it is perfectly legal to divide both sides of the equation by $x$. We are left with the simpler equation $x + 1 = 0$, which has solution $x = -1$.

Putting the two cases together, we see that both $x = 0$ and $x = -1$ are solutions. Notice that if you had divided both sides by $x$ without thinking in cases, you would have missed the $x = 0$ solution.

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I don't think there can be any such situation where multiplying both sides by $\frac 1x$ gives a redundant equation (where there is either no solution or an infinite amount of solutions) or a "defective equation," as long as $x \neq 0$. If you done such, the balance in the equation remains. This is why you have to apply the same operation on one side to the other. As an example, take the linear equation $x + 2 = 5$. When you solve it, you get $x = 3$. But if you were to start by multiplying both sides of the equation by $\frac 1x$, you'll see something like this: $$\begin{align} x + 2 & = 5 && \text{Given} \\ \frac 1x (x + 2) & = 5\left(\frac 1x\right) && \text{Multiply both sides by $\frac 1x$} \\ \frac{x+2}{x} & = \frac 5x && \text{Result of above operation} \\ x \cdot \frac{x + 2}{x} & = \frac 5x \cdot x && \text{Multiplying both sides by $x$ to cancel the denominator out} \\ x + 2 & = 5 && \text{We end up back where we started} \\ x & = 3 && \text{Solving for $x$} \end{align} $$

It did almost nothing to the original equation! Hence, multiplying both sides by $\frac 1x$ doesn't change the solutions-- as long as $x \neq 0$.

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  • $\begingroup$ Observe that a problem arises if you use this technique to solve $x(x+2) = 0$ without being careful about the possibility that $x = 0$. $\endgroup$ – Austin Mohr Jun 28 '17 at 1:27
  • $\begingroup$ Oh......and one of the solutions is $x = 0$! I originally had another approach to this, should I edit that one in? $\endgroup$ – Obinna Nwakwue Jun 28 '17 at 1:29

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