# How do you graph using the intercepts for 5x+7y=77?

Nov 3, 2015

Plot the points for the x and y intercepts then draw a line through those points.

#### Explanation:

Given the equation
$\textcolor{w h i t e}{\text{XXX}} 5 x + 7 y = 77$

The x-intercept is the value of $x$ when $y = 0$
$\textcolor{w h i t e}{\text{XXX}} 5 x + 7 \left(0\right) = 77$
$\textcolor{w h i t e}{\text{XXX}} \Rightarrow x = 15.4$

The y-intercept is the value of $y$ when $x = 0$
$\textcolor{w h i t e}{\text{XXX}} 5 \left(0\right) + 7 y = 77$
$\textcolor{w h i t e}{\text{XXX}} \Rightarrow y = 11$

So the intercept points are at $\left(15.4 , 0\right)$ and $\left(0 , 11\right)$

If you plot these two points and draw a straight line through them, your graph should look something like:
graph{((x-15.4)^2 +y^2-0.1)(x^2+(y-11)^2-0.1)(5x+7y-77)=0 [-6.41, 25.63, -3.47, 12.55]}