\(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(3x-\frac{3}{2}-5x-3=\frac{1}{5}-x\)
\(\left(3x-5x\right)-\left(\frac{3}{2}+3\right)=\frac{1}{5}-x\)
\(\left(-2\right).x-\frac{9}{2}=\frac{1}{5}-x\)
\(\left(-2\right).x+x.1=\frac{1}{5}+\frac{9}{2}\)
\(-1.x=\frac{47}{10}\)
\(x=\frac{47}{10}:\left(-1\right)\)
\(x=\frac{47}{10}.\frac{1}{-1}\)
\(x=\frac{47}{-10}\)
Vậy \(x=\frac{47}{-10}\)