How do you graph 3x - 2y ≥ 63x2y6?

1 Answer
Jul 1, 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 = 0x=0

(3 * 0) - 2y = 6(30)2y=6

0 - 2y = 602y=6

-2y = 62y=6

(-2y)/color(red)(-2) = 6/color(red)(-2)2y2=62

y = -3y=3 or (0, -3)(0,3)

For: y = 0y=0

3x - (2 xx 0) = 63x(2×0)=6

3x - 0 = 63x0=6

3x = 63x=6

(3x)/color(red)(3) = 6/color(red)(3)3x3=63

x = 2x=2 or (2, 0)(2,0)

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.
The boundary line will be solid because the inequality operator contains an "or equal to" clause.

graph{(x^2+(y+3)^2-0.04)((x-2)^2+y^2-0.04)(3x-2y-6)=0 [-10, 10, -5, 5]}

Now, we can shade the right side of the line.

graph{(3x-2y-6) >= 0 [-10, 10, -5, 5]}