Hello There!!
Given
[tex]3(x - 2) + 7x = \frac{1}{2} (6x - 2) [/tex]
Solving It
[tex]3(x - 2) + 7x = \frac{1}{2} (6x - 2) \\ \\ \implies3x - 6 + 7x = ([\frac{1}{2} \times 6 \: x] - [\frac{1}{2} \times 2]) \\ \\ \implies10x - 6 = ([\frac{1}{ \cancel2} \times \cancel 6 \: x] - [\frac{1}{ \cancel2} \times \cancel2]) \\ \\ \implies 10x - 6 = 3x - 1 \\ \\ \fbox{Bringing (3x )to left side whereas( - 6 )in right side} \\ \\ \implies10x - 3x = - 1 + 6 \\ \\ \implies7x = 5 \\ \\ \therefore x = \frac{5}{7} [/tex]
The Solution is = [tex] \frac{5}{7}[/tex]
[tex]\text\red{One Solution set is possible.}[/tex]
Hope This Helps
