Ta có: \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)
<=> \(12\left(\dfrac{3x-2}{6}-5\right)=12.\dfrac{3-2\left(x+7\right)}{4}\)
<=> \(6x-4-60=9-6\left(x+7\right)\)
<=> \(6x-64=9-6x-42\)
<=> \(6x-64=-6x-33\)
<=> \(6x+6x-64+33=0\)
<=> 12x-31=0
vậy \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)
<=> 12x-31=0