Ta có: \(4\left(3-x\right)-5\left(x-1\right)=x+20\)
\(\Leftrightarrow12-4x-5x+5-x-20=0\)
\(\Leftrightarrow-10x=3\)
hay \(x=-\dfrac{3}{10}\)
\(4\cdot\left(3-x\right)-5\cdot\left(x-1\right)=x+20\)
\(\Leftrightarrow-4x-5x-x=-12-5+20\)
\(\Leftrightarrow-10x=3\)
\(\Leftrightarrow x=-\dfrac{3}{10}\)