Calculation:

Consider, the equation, \(\displaystyle{4}{y}—{0.75}{x}+{1.2}={0}\),

To compute x-intercept, put \(\displaystyle{y}={0}\),

\(\displaystyle{4}\cdot{0}-{0.75}{x}+{1.2}={0}\)

\(\displaystyle{0.75}{x}+{1.2}-{1.2}={0}-{1.2}\)

\(\displaystyle{\frac{{{0.75}}}{{{0.75}}}}{x}={\frac{{{1.2}}}{{{0.75}}}}\)

\(\displaystyle{x}={1.6}\)

So, the x-intercept is \(\displaystyle{\left({1.6},{0}\right)}\).

To compute y -intercept, put \(\displaystyle{x}={0}\),

\(\displaystyle{4}{y}-{0.75}\cdot{x}+{1.2}={0}\)

\(\displaystyle{4}{y}+{1.2}-{1.2}={0}-{1.2}\)

\(\displaystyle{\frac{{{4}{y}}}{{{4}}}}={\frac{{-{1.2}}}{{{4}}}}\)

\(\displaystyle{y}=-{0.3}\)

So, the y-intercept is \(\displaystyle{\left({0},-{0.3}\right)}\).

Hence, thex and y -intercepts of \(\displaystyle{4}{y}—{0.75}{x}+{1.2}={0}\) are \(\displaystyle{\left({1.6},{0}\right)}\) and \(\displaystyle{\left({0},-{0.3}\right)}\), respectively.

Consider, the equation, \(\displaystyle{4}{y}—{0.75}{x}+{1.2}={0}\),

To compute x-intercept, put \(\displaystyle{y}={0}\),

\(\displaystyle{4}\cdot{0}-{0.75}{x}+{1.2}={0}\)

\(\displaystyle{0.75}{x}+{1.2}-{1.2}={0}-{1.2}\)

\(\displaystyle{\frac{{{0.75}}}{{{0.75}}}}{x}={\frac{{{1.2}}}{{{0.75}}}}\)

\(\displaystyle{x}={1.6}\)

So, the x-intercept is \(\displaystyle{\left({1.6},{0}\right)}\).

To compute y -intercept, put \(\displaystyle{x}={0}\),

\(\displaystyle{4}{y}-{0.75}\cdot{x}+{1.2}={0}\)

\(\displaystyle{4}{y}+{1.2}-{1.2}={0}-{1.2}\)

\(\displaystyle{\frac{{{4}{y}}}{{{4}}}}={\frac{{-{1.2}}}{{{4}}}}\)

\(\displaystyle{y}=-{0.3}\)

So, the y-intercept is \(\displaystyle{\left({0},-{0.3}\right)}\).

Hence, thex and y -intercepts of \(\displaystyle{4}{y}—{0.75}{x}+{1.2}={0}\) are \(\displaystyle{\left({1.6},{0}\right)}\) and \(\displaystyle{\left({0},-{0.3}\right)}\), respectively.