We are given with an equation and need to find it's solution set also we are given with a condition that m ≠ {0, 1} . So , let's start ;
So we are given with :
[tex]{:\implies \quad \sf \dfrac{6}{m}+2=\dfrac{m+3}{m-1}}[/tex]
First , we will simplify the given Equation and then we will solve it . So , now taking LCM in LHS ;
[tex]{:\implies \quad \sf \dfrac{6+2m}{m}=\dfrac{m+3}{m-1}}[/tex]
Now , Cross Multiplying ;
[tex]{:\implies \quad \sf (m-1)(6+2m)=m(m+3)}[/tex]
[tex]{:\implies \quad \sf m(6+2m)-1(6+2m)=m^{2}+3m}[/tex]
[tex]{:\implies \quad \sf 6m+2m^{2}-6-2m=m^{2}+3m}[/tex]
Now , transposing whole RHS to LHS and collecting like terms ;
[tex]{:\implies \quad \sf (2m^{2}-m^{2})+(6m-2m-3m)-6=0}[/tex]
[tex]{:\implies \quad \sf m^{2}+m-6=0}[/tex]
Now , we will solve this by splitting the Middle term ;
[tex]{:\implies \quad \sf m^{2}+3m-2m-6=0}[/tex]
[tex]{:\implies \quad \sf m(m+3)-2(m+3)=0}[/tex]
[tex]{:\implies \quad \sf (m+3)(m-2)=0}[/tex]
So , now either , m+3 = 0 or m-2 = 0 . On simplification we will get , m = 2 , -3. So m = {-3, 2}
Hence Option 1) {-3, 2} is correct
