How do you graph 4x=2y+6 using the x and y intercepts?

Nov 9, 2017

See a solution process below:

Explanation:

First, we need to determine the $x$ and $y$ intercepts.

x-intercept

Set $y$ equal to $0$ and solve for $x$:

$4 x = \left(2 \cdot 0\right) + 6$

$4 x = 0 + 6$

$4 x = 6$

$\frac{4 x}{\textcolor{red}{4}} = \frac{6}{\textcolor{red}{4}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} x}{\cancel{\textcolor{red}{4}}} = \frac{3}{2}$

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

$\left(\frac{3}{2} , 0\right)$

x-intercept

Set $x$ equal to $0$ and solve for $y$:

$4 \cdot 0 = 2 y + 6$

$0 = 2 y + 6$

$0 - \textcolor{red}{6} = 2 y + 6 - \textcolor{red}{6}$

$- 6 = 2 y + 0$

$- 6 = 2 y$

$- \frac{6}{\textcolor{red}{2}} = \frac{2 y}{\textcolor{red}{2}}$

$- 3 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} y}{\cancel{\textcolor{red}{2}}}$

$- 3 = y$

$\left(0 , - 3\right)$

We can next plot the two points on the coordinate plane:

graph{(x^2+(y+3)^2-0.025)((x-(3/2))^2+y^2-0.025)=0 [-10, 10, -5, 5]}

Now, we can draw a straight line through the two points to graph the line:

graph{(4x-2y-6)(x^2+(y+3)^2-0.025)((x-(3/2))^2+y^2-0.025)=0 [-10, 10, -5, 5]}