When am I allowed to remove a term by dividing 0 by the term

Question

It seems that you are allowed to remove terms from an equation if there is a 0 on the other side of the equals sign: $$ 3x = 0 $$ $$ \frac{3x}{3} = \frac{0}{3} $$ $$ x = 0 $$ However, you can't just do whatever you want: $$ 3x = 0 $$ $$ \frac{3x}{x} = \frac{0}{x} $$ $$ 3 \ne 0 $$

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$$ 3x^2 + 2x - 4 = 0 $$ $$ \frac{3x^2 + 2x - 4}{3x^2 + 2x - 4} = \frac{0}{3x^2 + 2x - 4} $$ $$ 1 \ne 0 $$ What exactly is the rule here? Is it simply unsafe only when the denominator would contain a variable or be 0? When is dividing 0 by something to remove that something from the equation allowed/safe, if ever?

Answer 1

You're dividing by zero in both cases which leads to fallacious results.

In the first equation, solving for $x$ leads to $x=0$. Thus, you cannot divide by it. You can divide by $3$ because $3 \ne 0$.

The same applies for the latter. You are explicitly guaranteeing it is zero because you are equating it to $0$.

Answer 2

When you work in a field as $\mathbb{R}$ or $\mathbb{C}$, dividing by $0$ is not allowed. On the other hand, you can divide by any non-zero element.

Consider $3x = 0$ in $\mathbb{R}$. If you multiply both sides by $\frac{1}{3}$ you get $x = 0$. So you cannot divide by $x$.

In the last case, you have $3x^2+2x-4=0$. This forces $x$ to be one of the two real roots of the polynomial in question. So you cannot divide both sides by the polynomial, because you would divide by $0$.

Answer 3

You can't ever divide by something that is zero. And if something might be zero, you must consider that it might not be to divide by the thing you are dividing.

Suppose you had $x^2 +x -6 = (x+2)(x-3) = 0$

But you are allowed to try. If you do try and get a valid result you can state the result is valid but only if the thing you were dividing by was not $0$.

So I can say if $x-3 \ne 0$ then I can do the following

$\frac {(x+2)(x-3)}{x-3}=\frac {0}{x-3} $ (!!!IF!!! $x -3 \ne 0$)

$x+2 = 0$

$x=-2$.

So I can conclude $x = 2$ IF $x -3 \ne 0$.

But if I do that I MUST immediately also consider the case that will happen if $x-3$ DOES equal $0$.

So I must conclude.

$x = -2$ if $x-3 \ne 0$ and if $x -3$ does equal $0$ then $x = 3$.

So either $x = -2$ or $x = 3$.

However if I try to divide and don't get a valid result, then we can conlcude we weren't allowed to do that.

Example: Let $(x^2 + 2)(x-3) = 0$

And I try: IF $x - 3 \ne 0$ then

$\frac {(x^2 + 2)(x-3)}{x-3} =\frac 0{x-3}$

$x^2 + 2 = 0$

so $x^2 = -2$. .... but that's impossible.

Then I must conclude, hmmm, I guess $x - 3\ne 0$ was false, and instead $x -3 = 0$ so $x = 3$.

Note: that is exactly the result we would have gotten if we had divide by $x^2 + 2$ instead.

Ex: Now we know $x^2 \ge 0$ so $x^2 + 2 \ge 2$ so $x^2 + 2 \ne 0$. So we can do

$\frac {(x^2 +2)(x-3)}{x^2 + 2} = \frac 0{x^2 + 2}$

$x -3 = 0$

$x = 3$.

Now one nifty thing about $a*b = 0$.

If you have that, then you know without a doubt, that one of $a$ or $b$ (or possibly both) is equal to $0$.

Answer 4

Rule #1

You can never divide a number or an expression by $0$

Rule #2

If you are dividing a number or an expression by an expression make sure that the denominator is not zero. For example $$3x=0 \iff \frac {3}{x}={0}{x}$$ only if $x\ne 0$

Rule #3

Do not multiply both sides of an equation by another expression without making sure that the expression is not $0$

for example $$ \frac {3x+2}{x-5}= \frac {x+4}{x-5} \iff {3x+2}={x+4}$$ only if $x\ne 5 $