Question 1043006
<pre><b>Well, in this case substituting k = x = h = 1
did happen to work, since
 
{{{(2k)/(2x+h)=(k)/(x+h)}}}

{{{(2*1)/(2*1+1)=(1)/(1+1)}}}

{{{2/3=1/2}}}

which is false.  So k=x=h=1 does produce
a counter-example.

However if the problem had been instead

{{{(2k)/(3x+h)=(k+4)/(6x+4h)}}}

and you substituted k = x = h = 1 

{{{(2*1)/(3*1+1)=(1+4)/(6*1+4*1)}}}

{{{2/4=5/10}}}

{{{1/2=1/2}}}

That would have been true, not false.  So you would
in that case need to use another number besides 1,
say 2, in order to get a counter-example.

For instance if you substituted k = x = h = 2 in

{{{(2k)/(3x+h)=(k+4)/(6x+4h)}}}

you get

{{{(4)/(8)=(6)/(20)}}}

{{{1/2=3/10}}}

which is false.

My point is that you may have to substitute other
values for x, h, and k besides 1 in order to find a 
counter-example.  Never assume that if you get a true 
equation, that the given equation is necessarily true.

In fact in some cases you cannot use the same value
for all the variables.  Take

{{{(k^2h)/(x+k)}}}{{{""=""}}}{{{khx/(x+h)}}}

If you use the same values for all three variables
you will never get a counter-example.

You'd have to use something like k=1, h=2, and k=3
to get a counter-example.

Edwin</pre></b>