SOLUTION: how would I solve {{{(x-4)/(4-x)}}}?

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Question 393696: how would I solve %28x-4%29%2F%284-x%29?
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
There's nothing to solve.
It's not an equation, there's no equal sign.
---------------
It can be simplified.
(x-4)/(4-x) = -1

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!


Many students say "solve" when they mean "simplify".  You say "solve an
equation or inequality" but when there is no equation or inequality you
have only an expression, so you can only say "simplify" or "rewrite",
not "solve".  Your problem has no equal sign or inequality sign, so you
can only say "simplify this":  

x - 4
-----
4 - x

The numerator is in standard order but the denominator is not.
Let's write the denominator in standard order too.

 x - 4
-------
-x + 4

Now the denominator is in the same order as the numerator, that is,
a variable term and then a numerical term.

Notice that the coefficient of x in the numerator is 1 and the
coefficient of x in the denominator is -1. Let's put
those in:

 1x - 4
--------
-1x + 4

Now you see that you can factor out 1 in the numerator and -1 in
the denominator, like this:

 1(x - 4)
----------
-1(x - 4)

Notice that when I factored out the -1 in the denominator that 
the +4 because a -4 inside the parentheses on the bottom.

Now the (x - 4)'s will cancel 

 1(x - 4)
----------
-1(x - 4)

and all that's left is

   1
 ----
  -1

which is just -1.

So the answer is -1.

--------------------------------------------------

A shortcut rule is to realize that each of these

(A - B) and (B - A)

is -1 times the other.  However this is a dangerous rule to 
learn unless you learn it very precisely and don't get careless
and think (A - B) and (A + B) can be handled by this rule. 

However if you knew that rule precisely you could just say

x - 4
-----  
4 - x
 
and in two steps you'd have 

-1
x - 4
-----
4 - x


-1.

Edwin