Solving equations with parameter m Discuss by the value of parameter m the solutions of these equations : 
$$    1). (m-1)x + 2m = 1 = 0$$  The first one is false 
$$     2). (2-m)x+3mx+2(m-x)-6=0 $$         I only did half the steps : Expanding , I found the value of x but with m      which is    $$ x =  \frac 3m  -1  $$    (That's the answer I found ) 
   I don't know if it's right and whether I should continue like this but at least I tried solving it first alone 
$$     3). (3m+5)x+3m = (2m-5)x+m+1 $$
 A: For the first equation It's False but then I thought I could change it right and now it became :    $$ (m-1)x+2m-1 = 0 $$ 
                        $$  (m-1)x = 1-2m    $$
     If   $$ m= 1  $$     Then  $$ 0x = -1 $$
                          $$  S= \varnothing   $$
If $$ m \neq 1  $$            Then  $$ x = \frac{1-2m}{m-1} $$
                           $$  S = \{\frac{1-2m}{m-1}\}$$ 
The second equation :        $$ (2-m)x + 3mx+2(m-x)-6=0 $$
                              $$ 2x-mx+3mx+2m-2x= 6  $$ 
                                $$ 2mx = 6-2m  $$
                                $$ mx = 3-m $$ 
 If $$ m= 0 $$ 
                 $$ 0x = 3 $$ 
                  $$ S = \varnothing   $$  
If $$ m\neq 0 $$
                   $$ x =  \frac{3-m}{m} $$
                    $$   S = \{\frac{3-m}{m}\}$$ 
The third equation :
               $$  (3m+5)x+3m = (2m-5)x+m+1 $$
              $$ ( 3m+5)x -(2m-5)x = -2m+1 $$
                $$ x (m+10) = -2m+1 $$
If $$ m=-10 $$ 
  Then $$ 0x =21 $$ 
                  $$ S =  \varnothing   $$  
If $$ m \neq -10 $$ 
          Then    $$ x = \frac{-2m+1}{m+10} $$ 
                  $$  S = \{\frac{-2m+1}{m+10}\}$$ 
Thank you for your best comments and answers Tell me If I did anything wrong 
A: You are expected to solve your equations for $x$ considering $m$ is a parameter to be supplied.  As your first equation has two equal signs I will delete the $=0$ part.  So $$(m-1)x + 2m = 1 \\x=\frac {1-2m}{m-1}$$ and you are supposed to discuss the effect of values of $m$ on the solution.  You would note that if $m=1$ the equation becomes $2=1$, which is false.  Assuming you are working in the reals, for any other value of $m$ there is a unique value of $x$.
A: int the equation $$x=\frac{3}{m}-1$$ should be $$m\ne 0$$. From the third equation we get $$(3m+5)x-(2m-5)x=-2m+1$$ thus we have $$x(m+10)=-2m+1$$ if $$m=-10$$ we obtain $$0=21$$ which is impossible. For $$m\ne -10$$ we get $$x=\frac{-2m+1}{m+10}$$
