When is it allowed to multiply with zero in algebra Consider the example:
$$\frac{7}{x} = 0$$
If we multiply both sides by $x$ we get
$$7 = 0$$
which obviously is wrong.
However with the equation
$$\frac{7x^2 - 67x +126}{x(x-2)(x-3)} = 0$$
You can multiply by zero and get
$$7x^2 -67x + 126 = 0$$
I think it is because there are still $x$'s remaining in the expression. However what is the exact rule when it is allowed to multiply by zero and when it is not...
 A: An equation of the form $$\frac{f(x)}{x} =g(x)$$ can only make sense when $x\ne 0$. You’re free to multiply with $x$ here. However, for an equation of the form $$f(x)=g(x) $$ You can only divide by $x$ if you know that $x\ne 0$. Note that there is no restriction on multiplying here, too. As for your first equation, $$\frac 7x=0$$ has no solutions.
A: You can multiply $0$ both side of an eauality anytime. But here nowhere we are multiplying $0$ both side.
Both are equations in $x$. For the first one if we assume there is some nonzero $x$ satisfying $\dfrac 7x=0$ then we arrive at a contradiction so there is no such $x$.
For the second one again let there exists $x$ such that $x(x-2)(x-3)$ is nonzero and $$\dfrac{7x^2-67x+126}{x(x-2)(x-3)}=0.$$ Then multiplying $x(x-2)(x-3)$ both side we get $7x^2-67x+126=0$.
A: In your first case, $$\frac 7 x = 0$$ doesn't have a solution. This is because $$\frac yx$$ is only equal to 0 if $$y = 0$$ If x=0 the expression is undefined.
In your second case we can deduce that, as the fraction is equal to 0, the numerator must be equal to 0. This gives us $$7x^2 - 67x + 126 = 0$$ which can then be solved.
A: Though your question shows something different, you can always "multiply by zero" the members of an equation, provided they have a defined value.

*

*$\dfrac 7x=0\implies0=0$ provided $x\ne0$ (otherwise the LHS is not defined).


*$\dfrac{7x^2 - 67x +126}{x(x-2)(x-3)} = 0\implies0=0$ provided $x\notin\{0,2,3\}$ (otherwise the LHS is not defined).

What you should have asked instead is "is it allowed to multiply by nonzero in algebra", and the answer is also yes (provided the members have a defined value).

*

*$\dfrac 7x=0\land x\ne0\implies 7=0$ proves that there is no suitable $x$.


*$\dfrac{7x^2 - 67x +126}{x(x-2)(x-3)} = 0\land x\notin\{0,2,3\}\implies 7x^2 - 67x +126=0\\\implies x=\dfrac{18}7\lor x=7$.
A: In your first example there is no solution to equation, and in you second equation there can be a solution. There is no rule, you can do that anytime you want.
