# Find the x-and y-intercepts of the graph of the equation algebraically. y=4y-0.75x+1.2=0

Find the x-and y-intercepts of the graph of the equation algebraically.
$$\displaystyle{y}={4}{y}-{0.75}{x}+{1.2}={0}$$

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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.
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