$$\frac{ 3 }{ 5 } (x-4)- \frac{ 1 }{ 3 } (2x-9)= \frac{ 1 }{ 4 } (x-1)-2$$
$x=9$
$$\frac{3}{5}x+\frac{3}{5}\left(-4\right)-\frac{1}{3}\left(2x-9\right)=\frac{1}{4}\left(x-1\right)-2$$
$$\frac{3}{5}x+\frac{3\left(-4\right)}{5}-\frac{1}{3}\left(2x-9\right)=\frac{1}{4}\left(x-1\right)-2$$
$$\frac{3}{5}x+\frac{-12}{5}-\frac{1}{3}\left(2x-9\right)=\frac{1}{4}\left(x-1\right)-2$$
$$\frac{3}{5}x-\frac{12}{5}-\frac{1}{3}\left(2x-9\right)=\frac{1}{4}\left(x-1\right)-2$$
$$\frac{3}{5}x-\frac{12}{5}-\frac{1}{3}\times 2x-\frac{1}{3}\left(-9\right)=\frac{1}{4}\left(x-1\right)-2$$
$$\frac{3}{5}x-\frac{12}{5}+\frac{-2}{3}x-\frac{1}{3}\left(-9\right)=\frac{1}{4}\left(x-1\right)-2$$
$$\frac{3}{5}x-\frac{12}{5}-\frac{2}{3}x-\frac{1}{3}\left(-9\right)=\frac{1}{4}\left(x-1\right)-2$$
$$\frac{3}{5}x-\frac{12}{5}-\frac{2}{3}x+\frac{-\left(-9\right)}{3}=\frac{1}{4}\left(x-1\right)-2$$
$$\frac{3}{5}x-\frac{12}{5}-\frac{2}{3}x+\frac{9}{3}=\frac{1}{4}\left(x-1\right)-2$$
$$\frac{3}{5}x-\frac{12}{5}-\frac{2}{3}x+3=\frac{1}{4}\left(x-1\right)-2$$
$$-\frac{1}{15}x-\frac{12}{5}+3=\frac{1}{4}\left(x-1\right)-2$$
$$-\frac{1}{15}x-\frac{12}{5}+\frac{15}{5}=\frac{1}{4}\left(x-1\right)-2$$
$$-\frac{1}{15}x+\frac{-12+15}{5}=\frac{1}{4}\left(x-1\right)-2$$
$$-\frac{1}{15}x+\frac{3}{5}=\frac{1}{4}\left(x-1\right)-2$$
$$-\frac{1}{15}x+\frac{3}{5}=\frac{1}{4}x+\frac{1}{4}\left(-1\right)-2$$
$$-\frac{1}{15}x+\frac{3}{5}=\frac{1}{4}x-\frac{1}{4}-2$$
$$-\frac{1}{15}x+\frac{3}{5}=\frac{1}{4}x-\frac{1}{4}-\frac{8}{4}$$
$$-\frac{1}{15}x+\frac{3}{5}=\frac{1}{4}x+\frac{-1-8}{4}$$
$$-\frac{1}{15}x+\frac{3}{5}=\frac{1}{4}x-\frac{9}{4}$$
$$-\frac{1}{15}x+\frac{3}{5}-\frac{1}{4}x=-\frac{9}{4}$$
$$-\frac{19}{60}x+\frac{3}{5}=-\frac{9}{4}$$
$$-\frac{19}{60}x=-\frac{9}{4}-\frac{3}{5}$$
$$-\frac{19}{60}x=-\frac{45}{20}-\frac{12}{20}$$
$$-\frac{19}{60}x=\frac{-45-12}{20}$$
$$-\frac{19}{60}x=-\frac{57}{20}$$
$$x=-\frac{57}{20}\left(-\frac{60}{19}\right)$$
$$x=\frac{-57\left(-60\right)}{20\times 19}$$
$$x=\frac{3420}{380}$$
$$x=9$$
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