# How do you graph using slope and intercept of 3/2x - 1/3y = 7/3?

Dec 31, 2015

Manipulate so that the equation is in the form $y = m x + b$, where $m$ is the slope and $b$ is the $y$-intercept.

First, to isolate the $y$-term, subtract $\frac{3}{2} x$ from both sides.

$- \frac{1}{3} y = - \frac{3}{2} x + \frac{7}{3}$

To isolate just $y$, multiply both sides by $- 3$.

$y = \frac{9}{2} x - 7$

This means that the function has an intercept at $\left(0 , - 7\right)$.

The slope is $\frac{9}{2}$. Slope is also known as "rise over run", since it represents the change in $y$-values over the change in $x$-values.

Thus, the "rise" is $9$ for every "run" of $2$. This means that if you have the point $\left(0 , - 7\right)$, you know that the point $\left(0 + 2 , - 7 + 9\right) \implies \left(2 , 2\right)$ is also on the line.

Graph $\left(2 , 2\right)$ and draw the line. If you wish, you can continue going up $9$ units and right $2$ units to draw in more points. You can also go down $9$ units and left $2$ units.

graph{3/2x-1/3y=7/3 [-28.47, 44.6, -15.8, 20.72]}