Why is $\frac{1}{4/3} - \frac{1}{3/2}$ not the same as $\bigl(\frac{4}{3} - \frac{3}{2}\bigr)^{-1}$ If you have the problem:$$\frac{1}{\frac{4}{3}} - \frac{1}{\frac{3}{2}} =?$$
Why can't you change the problem to $(\frac{4}{3} - \frac{3}{2})^{-1}$   and get the same answer?
In the first scenario, the answer is $1/12$
In the second scenario, the answer is $-6$
Why doesn't this work?
 A: You are basically asking why $\frac{1}{a} + \frac{1}{b} \neq \frac{1}{a+b}$. Take $ a = b = 2$, for example. You would expect that adding a half plus a half would get you back to one, i.e. $\frac{1}{2}+\frac{1}{2} = 1$, and certainly not a quarter, $\frac{1}{2+2} = \frac{1}{4}$.
Let's say you have some function $f(x)$ that takes some number $x$ and spits out a different number $y$. Let's take $f(x) = 3\cdot x$. If we put in a number that is a sum of two other numbers, for example, $x = a+b$, then we have that $f(x) = f(a+b) = f(a) + f(b)$. Same if $x$ is some multiple of a different number, $x = 4\cdot a$, namely $f(x) = f(4\cdot a) = 4f(a)$. These properties only hold because we choose our $f(x)$ to be very specific, i.e. multiplication by $3$. In general they do not have to hold however! For example if $f(x) = x^2$, or $f(x) = \frac{1}{x}$.
You are probably too used to functions behaving in the way $f(x) = 3\cdot x$ did, but if you look carefully almost everything that you can come up with will not have these nice properties.
A: When calculating this answer you have to honor the order of operations. The error here is that $$(a+b)^c\ne a^c+b^c.$$  you have to perform addition and then apply the exponent. These two operations are not commutative. In other words, it is not the same to apply the exponent to each value before adding. 
Hope this helps! 
A: The original problem is $$\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}$$
for specific values of $x,y$.  The modified problem is $$\frac{1}{x-y}$$
In order for them to be equal, $$\frac{y-x}{xy}=\frac{1}{x-y}$$
cross-multiplying, we get $$-(x-y)^2=xy$$
Expanding, we get $$-x^2+2xy-y^2=xy$$
or $$x^2-xy+y^2=0$$
The only real solutions this has are $x=0, y=0$, which are excluded since we are dividing by $x,y$.  Hence this proposed simplification is never correct, even accidentally.
A: Because $$(A-B)^{-1}\not=A^{-1}-B^{-1}.$$
Note that we have$$(\color{red}{A-B})^{-1}=\frac{1}{\color{red}{A-B}}.$$
For your case, we have
$$\left(\dfrac{4}{3}-\dfrac{3}{2}\right)^{-1}=\dfrac{1}{\dfrac{4}{3}-\dfrac{3}{2}}$$
A: It looks like based on your post that you want to know why $\frac{1}{\frac{4}{3}} - \frac{1}{\frac{3}{2}}$ is not equal to $\frac{1}{\frac{4}{3} - \frac{3}{2}}$.
In general, you can only say a property holds like, in your case, $(a + b)^{-1} = a^{-1} + b^{-1}$ if you can show it holds for every $a$ and $b$.  But as you pointed out in your question, if $a = \frac{4}{3}$ and $b = \frac{3}{2}$, this property doesn't hold.  Therefore, it's not true that $(a + b)^{-1} = a^{-1} + b^{-1}$ holds.  The property is false, as your example perfectly showed.
A: 1/(4/3 - 3/2) is not the same as 1/4/3 -1/3/2 simply because the latter expression is equivalent to 3/4 - 2/3 which is 9-8 / 12 which 12^-1. 12 is not 4/3 - 3/2.   In general,  fractions can only be grouped together by adding NUMERATORS if the DOMINATORS are the same. Unless they are being divided by each other or multiplied.  The reason for this can be shown from first principles but I think it's common sense tbh. 
