[tex]\huge{\boxed{\tt{Solution:-}}}[/tex]
[tex]\sf{(a.)\:x=\dfrac {-1}{2}}[/tex]
[tex]\sf{(b.)\:x=-2}[/tex]
[tex]\sf{(c.)\:x=\dfrac {1}{2}}[/tex]
[tex]\sf{(d.)\:x=2}[/tex]
Now, we solving it :
[tex]\sf{: \implies {4x - 6 = 10x - 3}} \\ \\ \sf{: \implies { - 6 + 3= 10x - 4x}} \\ \\ \sf{: \implies { - 3= 6x}} \\ \\ \sf{: \implies { \dfrac{- 3}{6} = x}} \\ \\ \sf{: \implies {\dfrac{- 1}{2} = x}} \\ \\ : \implies {\boxed{ \sf{x =\dfrac{- 1}{2}}}}[/tex]
[tex]\bf{\therefore{Correct \:solution\:is\:(a.)\:x=\dfrac {-1}{2}.}}[/tex]
[tex]\rule {307}{2}[/tex]
