# How do you graph the inequality 3x + y >1?

##### 1 Answer
Oct 3, 2017

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: $x = 0$

$\left(3 \cdot 0\right) + y = 1$

$0 + y = 1$

$y = 1$ or $\left(0 , 1\right)$

For: $x = 2$

$\left(3 \cdot 2\right) + y = 1$

$6 + y = 1$

$- \textcolor{red}{6} + 6 + y = - \textcolor{red}{6} + 1$

$0 + y = - 5$

$y = - 5$ or $\left(2 , - 5\right)$

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-1)^2-0.125)((x-2)^2+(y+5)^2-0.125)(3x+y-1)=0 [-20, 20, -10, 10]}

Now, we can make the boundary line dashed because the inequality operator does not contain an "or equal to" clause.

We can also shade to the right side of the line because of the "greater than" clause.

graph{(3x+y-1) > 0 [-20, 20, -10, 10]}