Why can't you cancel both xs with 2x/3x?

Question

Consider that -

1. $x^3/x^2 = x $ as the exponent of the denominator is one higher than that of the numerator.

2. $x^2/x^2 = 1$

With $3x/2x$ it is not possible to cancel both $x$.

It seems like $3x/2x$ could be shown as $3x^1/2x^1$ meaning the $x$ should cancel like with $x^2/x^2 = 1$ leaving $3*1/2*1$

Please could someone explain to me the missing link as I cannot understand why this is not the case?

this is the fraction it came from:

$3x-x^2/2x+4$ I was told it can not be simplified further but I thought that the $3x /2x$ part could be turned to $3/2$ as the $x$ would cancel.

Answer 1

In such expressions

$$\frac{x^m}{x^n}$$

we can always cancel out terms without change its value but under the condition that $x\neq 0$, indeed

• $\frac{x}{x}=1$ for all $x\neq 0$

but

• $\frac{0}{0}$ is not defined.

Answer 2

What you should do is to check the denominator is not 0.

If the denominator is 0, we cannot define it.

If denominator is not 0, we can do reduction.

Answer 3

With $3x/2x$ it is not possible to cancel both $x.$

Why not? It's impossible only if $x=0.$ Provided $x$ is nonzero, you can simplify further.

As for $$3x-x^2/2x+4,$$ you may very well simplify further since this expression has no definite value whenever $x=0$ anyway. So we may assume (implicitly -- which is what's almost always done) without losing anything significant that $x\ne 0,$ and then simplify the middle term to become $-x/2.$