How would I graph the inequality 3x+2y<-6 in a coordinate plane?

Question

Do I shade a certain side !? AH I need help!

Answer 1

3x+2y<-6

2y < -6 -3x
y < -3x/2 -3
draw the line y = -3x/2 -3

shade the region under the line (line is excluded).

Answer 2

The first answer was correct. Since y is less than (-1.5x - 6), the y-values that satisfy the inequality must be below the line y = -1.5x - 6. An easy way to check, if this confuses you, is to pick a point (x,y) that is well below the line and see if it satisfies the inequality. For example, take x = 0, y = -100, which is certainly below the line. At (0,-100),

3x + 2y < -6 becomes 3(0) + 2(-100) < -6, or

-200 < -6. That is certainly true, so yes, it is the lower area that is shaded.

Answer 3

Consider the line 3x + 2y = - 6
Points on this line are:-
(0 , - 3) , (- 2 , 0)
Draw a dotted line thro` these points.
The required area is then a shaded area to the LEFT of this dotted line.

Answer 4

get the inequality in slope-intercept form
2y<-6 - 3x
y<-6/2 - 3x/2
y< -3x/2 - 3
so y-intercept is (0 , -3)
and slope is -3/2
for y< make dotted line and shade below

Answer 5

Once you do something flipping around you get...

-1.5x-3>y

So you graph the line -1.5x-3=y and the area above the line
is shaded in because all of those points are correct.

Answer 6

3x+2y<-6
2y<-3x-6
y<(-3/2)x-3
so sketch y=-1.5x-x and shade under it since y is smaller
btw beware Brent's answer! it's wrong! you shade under it! and you make the line dotted