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

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

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Answer 1 · 2015-04-03T21:27:02

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

$2 (y - 1) = 2 y - 2$ $2(y-1)=2y-2$
$3 (x + 1) = 3 x + 3$ $3(x+1)=3x+3$ .
So, the expression is equivalent to $2 y - 2 > 3 x + 3$ $2y-2>3x+3$
Add $2$ $2$ to both sides:
$2 y > 3 x + 5$ $2y>3x+5$
Divide by $2$ $2$ both sides:
$y > \frac{3}{2} x + \frac{5}{2}$ $y>3/2 x + 5/2$

Now, it's easy to plot the equation $y = \frac{3}{2} x + \frac{5}{2}$ $y=3/2 x+ 5/2$ , since it is a line. That line is the graph of the equation, which means the set of points
which realize $y = \frac{3}{2} x + \frac{5}{2}$ $y=3/2 x + 5/2$ , and you want the set of points composed of all the greater $y$ $y$ '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 $y = \frac{3}{2} x + \frac{5}{2}$ $y=3/2 x + 5/2$ is not to be considered

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

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