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

1 Answer
KillerBunny
Apr 3, 2015

First of all, let's manipulate the expression to isolate the #y# term:

  • #2(y-1)=2y-2#
  • #3(x+1)=3x+3#.
    So, the expression is equivalent to #2y-2>3x+3#
    Add #2# to both sides:
    #2y>3x+5#
    Divide by #2# both sides:
    #y>3/2 x + 5/2#

Now, it's easy to plot the equation #y=3/2 x+ 5/2#, since it is a line. That line is the graph of the equation, which means the set of points
which realize #y=3/2 x + 5/2#, and you want the set of points composed of all the greater #y#'s. This simply means that you must consider the area above the line to solve the equation, as you can see in the graph:

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

Note that the line itself is not part of the solution, because you have the strict inequality and so the set of points #y=3/2 x + 5/2# is not to be considered

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