How do you graph x - 2/3y = 1 ?

1 Answer
May 30, 2017

First, let's rewrite this as point-slope form or $y = m x + b$:

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

subtract $x$ on both sides

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

divide both sides by $- \frac{2}{3}$

$y = \left(1 - x\right) \div - \frac{2}{3}$

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

$y = \frac{2 \left(- x + 1\right)}{3}$

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

$y = - \frac{2}{3} x + \frac{2}{3}$
Now it's in point-slope form ( $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$ )

the reason I wanted it in this exact form is because it gives us valuable information: the $\textcolor{red}{s l o p e}$ and the $\textcolor{b l u e}{y - i n t e r c e p t}$

These are very helpful for graphing, buecause we really only need to points on the line to draw the rest of it. And we already have one, the $y$-intercept:

$y = \textcolor{red}{- \frac{2}{3}} x + \textcolor{b l u e}{\frac{2}{3}}$
The $y$-intercept is $\left(0 , \frac{2}{3}\right)$, and it's very simple to find the $x$-intercept. All we need to do is set $y$ equal to zero and solve for $x$:

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

$0 = - \frac{2}{3} x + \frac{2}{3}$

subtract $\frac{2}{3}$ on both sides

$- \frac{2}{3} = - \frac{2}{3} x$

divide by $- \frac{2}{3}$ on both sides

$- \frac{2}{3} \div - \frac{2}{3} = x$

$\cancel{-} \frac{\cancel{2}}{\cancel{3}} \times \cancel{-} \frac{\cancel{3}}{\cancel{2}} = x$

$x = 1$

So, now we have two pints: $\left(0 , \frac{2}{3}\right)$ and $\left(1 , 0\right)$. Now you just need to use a rule to draw a line between these points, or start at one of them and use the slope to find more points:

Just to check our work, let's graph our equation. If we did everything correctly, the line should pass through the points $\left(0 , \frac{2}{3}\right)$ and $\left(1 , 0\right)$.

graph{y=-(2)/3x + 2/3}

We were right!