Question

\(3\left(x-\dfrac{1}{3}\right)-7\left(x+\dfrac{3}{7}\right)=-2x+\dfrac{1}{3}\)

Answer 1

`3(x-1/3)-7(x+3/7)=-2x+1/3`

`3x-1-7x-3=-2x+1/3`

`3x-7x+2x=1/3+1+3`

`-2x=13/3`

`x=13/3:(-2)=[-13]/6`

Answer 2

\(3\left(x-\dfrac{1}{3}\right)-7\left(x+\dfrac{3}{7}\right)=-2x+\dfrac{1}{3}\)

\(3x+3\left(-\dfrac{1}{3}\right)-7\left(x+\dfrac{3}{7}\right)=-2x+\dfrac{1}{3}\)

\(3x-1-7\left(x+\dfrac{3}{7}\right)=-2x+\dfrac{1}{3}\)

\(3x-1-7x-7.\left(\dfrac{3}{7}\right)=-2x+\dfrac{1}{3}\)

\(3x-1-7x-3=-2x+\dfrac{1}{3}\)

\(-4x-1-3=-2x+\dfrac{1}{3}\)

\(-4x-4=-2x+\dfrac{1}{3}\)

\(-4x-4+2x=\dfrac{1}{3}\)

\(-2x-4=\dfrac{1}{3}\)

\(-2x=\dfrac{1}{3}+4\)

\(-2x=\dfrac{13}{3}\)

\(x=\dfrac{13}{3}:-2\)

\(x=-\dfrac{13}{6}\)

Answer 3

                     

Answer 4

\(3x-1-7x+3=-2x+\dfrac{1}{3}\)

\(3x-1-7x+3+2x-\dfrac{1}{3}=0\)

\(\Leftrightarrow-2x+\dfrac{5}{3}=0\)

\(=>-2x=-\dfrac{5}{3}\Rightarrow x=\dfrac{5}{3}:\left(-2\right)=-\dfrac{5}{6}\)