If $3ab^2 = 4$ and $3ab = -5$, can we say that $\frac{3ab^2}{3ab} = -\frac{4}{5}$?

Question

I was wondering if this is true:

If i have $3ab^2 = 4$ and $3ab = -5$

Can i divide both of them, so that $\frac{3ab^2}{3ab} = -\frac{4}{5},$ so that $b = -\frac{4}{5}?$

Answer 1

Yes upon checking you are not dividing by $0$. Since $3ab=-5$, it is non-zero and we can divide it.

$$\frac{3ab^2}{3ab}=\frac{4}{3ab}=\frac{4}{-5}$$

Hence $b=-\frac45$ and then you can solve for $a$.

Answer 2

Yes, you can, since you know that the number you are dividing by is non-zero.

If you feel uneasy about dividing equations, you can avoid dividing equations the following way:

$$4=3ab^2=(3ab)b=(-5)b$$

Answer 3

The reason you can do this is because, if two things are equal, if you do the same valid (and deterministic) operation to both of them, they will remain equal.

Therefore, since both of these things are equations, that means that both sides are equal to each other. Since $3ab = -5$, then I can use $3ab$ and $-5$ as replacements for each other. Any place I can use $3ab$, I can also use $-5$.

On the original equation, you had $3ab^2 = 4$. That means that both of these expressions are the same. They will remain the same as long as you do the same valid operations to both sides.

So, dividing $3ab^2$ by $3ab$ is a logically valid thing to do (since $3ab = -5$ it can't be division by zero). The equation will remain equal as long as I do the same thing to both sides. So I will divide both sides by $3ab$. However, my equations also tell me that $-5$ is a valid stand-in for $3ab$. Therefore, if I divide one side by $3ab$ and the other side by $-5$, I'm actually dividing by the same value!

So, since $3ab^2$ and $4$ are the same, after doing the same operation to both sides, they will still be the same, and therefore, can still have the equal sign between them.