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

2 Answers
May 13, 2015

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]}

May 13, 2015

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]}