# Find the x-and y-intercepts of the graph of the equation algebraically. y=2x+3y=10

Question
Upper level algebra
Find the x-and y-intercepts of the graph of the equation algebraically.
$$\displaystyle{y}={2}{x}+{3}{y}={10}$$

2020-12-03
Calculation:
Consider, the equation $$\displaystyle{2}{x}+{3}{y}={10}$$,
To compute x-intercept, put $$\displaystyle{y}={0}$$.
$$\displaystyle{2}{x}+{3}\cdot{0}={10}$$
$$\displaystyle{2}{x}={10}$$
Divide both sides of the equation,
$$\displaystyle{\frac{{{2}{x}}}{{{2}}}}={\frac{{{10}}}{{{2}}}}$$
$$\displaystyle{x}={5}$$
So, the x-intercept is $$\displaystyle{\left({5},{0}\right)}$$.
To compute y-intercept, put $$\displaystyle{x}={0}$$,
$$\displaystyle{2}\cdot{0}+{3}{y}={10}$$
$$\displaystyle{3}{y}={10}$$
Divide both sides of the equation by 3,
$$\displaystyle{\frac{{{3}{y}}}{{{3}}}}={\frac{{{10}}}{{{3}}}}$$
$$\displaystyle{y}={\frac{{{10}}}{{{3}}}}$$
So, the y -intercept is $$\displaystyle{\left({0},{\frac{{{10}}}{{{3}}}}\right)}$$.
Hence, the x and y -intercept of $$\displaystyle{2}{x}+{3}{y}={10}$$ are $$\displaystyle{\left({5},{0}\right)}$$ and $$\displaystyle{\left({0},{\frac{{{10}}}{{{3}}}}\right)}$$, respectively.

### Relevant Questions

Find the x-and y-intercepts of the graph of the equation algebraically.
$$\displaystyle{\frac{{{2}{x}}}{{{5}}}}+{8}-{3}{y}={0}$$
Find the x-and y-intercepts of the graph of the equation algebraically.
$$\displaystyle{3}{y}+{2.5}{x}-{3.4}={0}$$
Find the x-and y-intercepts of the graph of the equation algebraically.
$$\displaystyle{y}=-{3}{\left({2}{x}+{1}\right)}$$
Find the x-and y-intercepts of the graph of the equation algebraically.
$$\displaystyle{y}={16}-{3}{x}$$
Find the x-and y-intercepts of the graph of the equation algebraically.
$$\displaystyle{4}{x}-{5}{y}={12}$$
Find the x-and y-intercepts of the graph of the equation algebraically.
$$\displaystyle{y}={5}-{\left({6}-{x}\right)}$$
$$\displaystyle{y}=-{5}{x}+{6}$$
$$\displaystyle{\frac{{{8}{x}}}{{{3}}}}+{50}-{2}{y}={0}$$
$$\displaystyle{y}={4}{y}-{0.75}{x}+{1.2}={0}$$
$$\displaystyle{y}={\frac{{{3}}}{{{4}}}}{x}-{\frac{{{1}}}{{{4}}}}$$