# How do you graph y-2<3x?

Jan 3, 2018

See a solution process below:

#### Explanation:

First, solve for two points as an equation instead of an inequality to find the boundary line for the inequality.

For: $x = 0$

$y - 2 = 3 \times 0$

$y - 2 = 0$

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

$y - 0 = 2$

$y = 2$ or $\left(0 , 2\right)$

For: $x = 1$

$y - 2 = 3 \times 1$

$y - 2 = 3$

$y - 2 + \textcolor{red}{2} = 3 + \textcolor{red}{2}$

$y - 0 = 5$

$y = 5$ or $\left(1 , 5\right)$

We can now graph the two points on the coordinate plane and draw a line through the points to mark the boundary of the inequality.

graph{(x^2+(y-2)^2-0.05)((x-1)^2+(y-5)^2-0.05)(y-3x-2)=0 [-16, 16, -8, 8]}

Now, we can shade the right side of the line. The boundary line will be dashed because the inequality operator does not contain an "or equal to" clause.

graph{(y-3x-2)<0 [-16, 16, -8, 8]}