Calculation:
\(\displaystyle{x}\ +\ {3.9}={4.2}\)
Substract 3.9 on both sides,
\(\displaystyle{x}\ +\ {3.9}\ -\ {3.9}={4.2}\ -\ {3.9}\)

\(\displaystyle{3.9}\ -\ {3.9}\) Based on the definition, \(\displaystyle{b}{e}{g}\in{\left\lbrace{e}{q}{u}{a}{t}{i}{o}{n}\right\rbrace}{\frac{{{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left[{b}\right]}{\left\lbrace{r}\right\rbrace}+{3.9}\backslash-\ {3.9}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}}}{{+{0.0}}}}{e}{n}{d}{\left\lbrace{e}{q}{u}{a}{t}{i}{o}{n}\right\rbrace}\)

\(\displaystyle{3.9}\ -\ {3.9}={0}\)

\(\displaystyle{4.2}\ -\ {3.9}\) Based on the definition, \(\displaystyle{b}{e}{g}\in{\left\lbrace{e}{q}{u}{a}{t}{i}{o}{n}\right\rbrace}{\frac{{{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left[{b}\right]}{\left\lbrace{r}\right\rbrace}+{4.2}\backslash-\ {3.9}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}}}{{+{0.3}}}}{e}{n}{d}{\left\lbrace{e}{q}{u}{a}{t}{i}{o}{n}\right\rbrace}\)

\(\displaystyle{4.2}\ -\ {3.9}={0.3}\) Substitute \(\displaystyle{3.9}\ -\ {3.9}={0}\ \text{and}\ {4.2}\ -\ {3.9}={0.3}\ \text{in}\ {x}\ +\ {3.9}\ -\ {3.9}={4.2}\ -\ {3.9}\)

\(\displaystyle{x}\ +\ {0}={0.3}\)

\(\displaystyle{x}={0.3}\) Final solution: \(\displaystyle{x}={0.3}\)

\(\displaystyle{3.9}\ -\ {3.9}\) Based on the definition, \(\displaystyle{b}{e}{g}\in{\left\lbrace{e}{q}{u}{a}{t}{i}{o}{n}\right\rbrace}{\frac{{{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left[{b}\right]}{\left\lbrace{r}\right\rbrace}+{3.9}\backslash-\ {3.9}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}}}{{+{0.0}}}}{e}{n}{d}{\left\lbrace{e}{q}{u}{a}{t}{i}{o}{n}\right\rbrace}\)

\(\displaystyle{3.9}\ -\ {3.9}={0}\)

\(\displaystyle{4.2}\ -\ {3.9}\) Based on the definition, \(\displaystyle{b}{e}{g}\in{\left\lbrace{e}{q}{u}{a}{t}{i}{o}{n}\right\rbrace}{\frac{{{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left[{b}\right]}{\left\lbrace{r}\right\rbrace}+{4.2}\backslash-\ {3.9}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}}}{{+{0.3}}}}{e}{n}{d}{\left\lbrace{e}{q}{u}{a}{t}{i}{o}{n}\right\rbrace}\)

\(\displaystyle{4.2}\ -\ {3.9}={0.3}\) Substitute \(\displaystyle{3.9}\ -\ {3.9}={0}\ \text{and}\ {4.2}\ -\ {3.9}={0.3}\ \text{in}\ {x}\ +\ {3.9}\ -\ {3.9}={4.2}\ -\ {3.9}\)

\(\displaystyle{x}\ +\ {0}={0.3}\)

\(\displaystyle{x}={0.3}\) Final solution: \(\displaystyle{x}={0.3}\)