\(\displaystyle{f}:{R}\to{R}\)

f(x)=7x-3

Let \(\displaystyle{y}\in{R}\) and

\(\displaystyle{x}=\frac{{{y}+{3}}}{{7}}\)

\(\displaystyle{x}=\frac{{{y}+{3}}}{{7}}\in{R}\)

f(x)=7x-3

\(\displaystyle={7}{\left(\frac{{{y}+{3}}}{{7}}\right)}-{3}\)

=y+3-3

=y

Since f(x)=y, and \(\displaystyle{y}\in{R}\), f is onto

f(x)=7x-3

Let \(\displaystyle{y}\in{R}\) and

\(\displaystyle{x}=\frac{{{y}+{3}}}{{7}}\)

\(\displaystyle{x}=\frac{{{y}+{3}}}{{7}}\in{R}\)

f(x)=7x-3

\(\displaystyle={7}{\left(\frac{{{y}+{3}}}{{7}}\right)}-{3}\)

=y+3-3

=y

Since f(x)=y, and \(\displaystyle{y}\in{R}\), f is onto