Question 1149579
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Right and &nbsp;<U>CANONICAL</U> &nbsp;way to do it is as follows.



<pre>
Let "t" be new variable, t = 2x+1.


Then  x = {{{(t-1)/2}}}.


Express  6x+2  via  "t" :


    6x+2 = {{{6*((t-1)/2)+2}}} = 3*(t-1) + 2 = 3t - 3 + 2 = 3t-1.


Thus  f(t) = 3t-1.


Here "t" is  a "mute" / ("silent") variable.  <U>It can be replaced by ANY symbol in both sides simultaneously.</U>


For example, it can be replaced by x.  So


    f(x) = 3x-1.      <U>ANSWER</U>
</pre>

Done, completed and explained.



This my post is the &nbsp;<U>standard &nbsp;MANTRA</U> &nbsp;to use for solution to each and every similar problem.



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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;//  &nbsp;&nbsp;Mantra is called "mantra", &nbsp;because it &nbsp;<U>DOES &nbsp;NOT &nbsp;ALLOW</U> &nbsp;to change any word (!)