Answer:
[tex]\huge\boxed{x=24}[/tex]
Step-by-step explanation:
[tex]\dfrac{1}{4}x-2=-6+\dfrac{5}{12}x\qquad|\text{add 2 to both sides}\\\\\dfrac{1}{4}x-2+2=-6+2+\dfrac{5}{12}x\\\\\dfrac{1}{4}x=-4+\dfrac{5}{12}x\qquad|\text{subtract}\ \dfrac{5}{12}x\ \text{from both sides}\\\\\dfrac{1}{4}x-\dfrac{5}{12}x=-4+\dfrac{5}{12}x-\dfrac{5}{12}x\\\\\dfrac{1\cdot3}{4\cdot3}x-\dfrac{5}{12}x=-4\\\\\dfrac{3}{12}x-\dfrac{5}{12}x=-4\\\\\dfrac{3-5}{12}x=-4\\\\\dfrac{-2}{12}x=-4\\\\\dfrac{-2:2}{12:2}x=-4[/tex]
[tex]-\dfrac{1}{6}x=-4\qquad|\text{multiply both sides by (-6)}\\\\\left(-6\!\!\!\!\diagup\right)\cdot\left(-\dfrac{1}{6\!\!\!\!\diagup}\right)x=(-4)(-6)\\\\x=24[/tex]
