# Find the x-and y-intercepts of the graph of the equation algebraically. y=16-3x

Find the x-and y-intercepts of the graph of the equation algebraically.
$$\displaystyle{y}={16}-{3}{x}$$

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Khribechy
Calculation:
Consider, the equation, $$\displaystyle{y}={16}—{3}{x}$$,
To compute x-intercept, put $$\displaystyle{y}={0}$$.
$$\displaystyle{0}={16}-{3}{x}$$
Add —16,to both the sides, this gives,
$$\displaystyle{0}-{16}={16}-{3}{x}-{16}$$
$$\displaystyle-{16}=-{3}{x}$$
Divide both sides of the equation by -3,
$$\displaystyle{\frac{{-{16}}}{{-{3}}}}={\frac{{-{3}{x}}}{{-{3}}}}$$
$$\displaystyle{x}={\frac{{{16}}}{{{3}}}}$$
So, the x-intercept is $$\displaystyle{\left({\frac{{{16}}}{{{3}}}},{0}\right)}$$.
To compute y -intercept, put $$\displaystyle{x}={0}$$,
$$\displaystyle{y}={16}-{3}{\left({0}\right)}$$
$$\displaystyle{y}={16}$$
So, the y-intercept is(0,16).
Hence, the x and y intercepts of $$\displaystyle{y}={16}-{3}{x}$$ are $$\displaystyle{\left({\frac{{{16}}}{{{3}}}},{0}\right)}$$ and(0,16),respectively.
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