\(\displaystyle\frac{{{5}+{x}-{2}}}{{3}}=\frac{{{x}+{3}}}{{8}}\)

\(\displaystyle\frac{{{5}+{x}-{2}}}{{3}}=\frac{{{x}+{3}}}{{8}}\)

Multiply both sides by \(\displaystyle{3}\times{8}={24}\)

Note that 24 is a product of 3 and 8, Which are the denominators of the fractions in LHS and RHS

\(\displaystyle{24}\times{5}+{24} \times \frac{{{x}-{2}}}{{3}}={24}\times\frac{{{x}+{3}}}{{8}}\)

\(\displaystyle{120}+{8}{\left({x}-{2}\right)}={3}{\left({x}+{3}\right)}\)

\(\displaystyle{120}+{8}{x}-{16}={3}{x}+{9}\)

\(\displaystyle{8}{x}+{104}={3}{x}+{9}\)

Subtract 3x from both sides, To get

\(\displaystyle{5}{x}+{104}={9}\)

Subtract 104 from both sides, To get

\(\displaystyle{5}{x}={9}-{104}\)

\(\displaystyle{b}{z}=-{95}\)

Divide both sides by 5, To get

\(\displaystyle{x}=\frac{{-{95}}}{{5}}=-{19}\)

We get the finally result

\(\displaystyle{x}=-{19}\)