Question

How do you graph `2(y-1) > 3(x+1)`?

Answer 1

The inequality can be rewritten under the form

`y > 3/2 x + 5/2`

after the following operations:

`2 y - 2 > 3 x + 3` <=> `2 y > 3 x + 5` <=> `y > 3/2 x + 5/2`

The graph of the equation `y = 3/2 x + 5/2` is a straight line. The graph of the inequality `y > 3/2 x + 5/2` will be formed by all the points of the plane whose coordinates `y` have values higher than `3/2 x + 5/2`

Therefore, the graph of our inequality will be the semi-plane situated above the blue line defined by the equation `y = 3/2 x + 5/2`

graph{y > 3/2 x + 5/2 [-10, 10, -5, 5]}

Answer 2

Distribute the `2` and the `3` .

`2(y-1)>3(x+1)`

`2y-2>3x+3`

Convert the inequality to slope-intercept form while keeping the inequality.

Add `-2` to both sides.

`2y>3x+3+2` =

`2y>3x+5`

Divide both sides by `2`.

`(cancel(2)y)/cancel2>(3x)/(2)+(5)/(2)` =

`y>(3x)/2+(5)/2`

In order to graph `y>(3x)/2+(5)/2`, do the following:

Determine two points on the line by temporarily making `y=(3x)/2+(5)/2` .

If `x=1`; `y=((3)*(1))/2+5/2=3/2+5/2=cancel8^4/cancel2^1=4`
Point = `(1,4)`

If `x=3`; `y=((3)*(3))/2+5/2=9/2+5/2=cancel14^7/cancel2^1=7`
Point = `(3,7)`

Plot the two points and draw a dashed line -------- through the two points. This represents that the points on the line are not part of the inequality. Then shade in the area above the dashed line.

graph{y>(3x)/2+(5)/2 [-16.02, 16, -8.01, 8.01]}