Solve the given equation as shown, follow the steps below:

\(\displaystyle{\left|{5}{x}-{1}\right|}={\left|{3}-{4}{x}\right|}{\left[{r}{e}{c}{a}{l}{l}:\ {\left|{a}\right|}={\left|{b}\right|}\Rightarrow{a}=\pm{b}\right]}\)

\(\displaystyle\Rightarrow{5}{x}-{1}=\pm{\left({3}-{4}{x}\right)}\) [add 1 in both sides]

\(\displaystyle\Rightarrow{5}{x}={1}\pm{\left({3}-{4}{x}\right)}\)

Solve the two equations that occured.

\(\displaystyle{5}{x}={1}-{\left({3}-{4}{x}\right)}\)

\(\displaystyle\Rightarrow{5}{x}={1}-{3}+{4}{x}\)

\(\displaystyle\Rightarrow{5}{x}-{4}{x}=-{2}\)

\(\displaystyle\Rightarrow{x}=-{2}\)

\(\displaystyle{5}{x}={1}+{\left({3}-{4}{x}\right)}\)

\(\displaystyle\Rightarrow{5}{x}={1}+{3}-{4}{x}\)

\(\displaystyle\Rightarrow{5}{x}+{4}{x}={1}+{3}\)

\(\displaystyle\Rightarrow{9}{x}={4}\)

\(\displaystyle\Rightarrow{x}={\frac{{{9}}}{{{4}}}}\)

The solutions of this equation are the x=-2 and \(\displaystyle{x}={\frac{{{9}}}{{{4}}}}\)