Verified answer

[tex]\displaystyle\bf\\1)\\\\\frac{2a-3}{x} -\frac{4a}{x} =\frac{2a-3-4a}{x} =-\frac{2a+3}{x} \\\\\\2)\\\\\frac{3c+5}{1-3c}+\frac{4c+6}{3c-1} = \frac{3c+5}{1-3c}-\frac{4c+6}{1-3c} = \frac{3c+5-4c-6}{1-3c}=\\\\\\=\frac{-c-1}{1-3c} =\frac{c+1}{3c-1}[/tex]