[tex]\frac{x-\frac{1}{3}}{\frac{1}{4}}-\frac{1}{12}=\frac{x+\frac{1}{3}}{\frac{4}{3}} \\
\\
\frac{\frac{3x-1}{3}}{\frac{1}{4}}-\frac{1}{12}=\frac{\frac{3x+1}{3}}{\frac{4}{3}} \\
\\
\frac{3x-1}{3}.4-\frac{1}{12}=\frac{3x+1}{3}.\frac{3}{4} \\
\\
\frac{12x-4}{3}-\frac{1}{12}=\frac{3x+1}{4} \\
\\
4(12x-4)-1=3(3x+1) \\
\\
48x-16-1=9x+3 \\
\\
48x-9x=3+16+1 \\
\\
39x=20 \\
\boxed{x=\frac{20}{39}}[/tex]
