Our system has two equations: y > -2 and x + y < 4.
A graph of y > -2 is a horizontal dashed line and everything above it is shaded. A graph of x + y < 4 shades everything to the left of it. From the graphs of this system you can determine an ordered pair that is true (so it is a solution) and an ordered pair that is false (not a solution).
A good test point - without a graph - is (0,0).
0 > -2 TRUE
0 + 0 = 0 and 0 < 4 TRUE
So (0, 0) is a solution in the system because it makes both equations true.
2. Now we want to find where the system is false. With a graph, it's the area unshaded by both. With the equations, we proceed as follows.
x + y < -4
y < - 4 - x Subtract x on both sides
Since y > -2 Then -2 < y. So by transitivity (if a < b and b < c, a < c), we have that
-2 < -4 - x
2 < -x add two to both sides
x > -2 multiply by -1 and change orientation.
So when x > -2 and y > -2 we have true inequalities - something we used above and found with (0,0). If we flip this around, we have false inequalities when x ≤ -2 or y ≤ -2. (The negation of an 'and' is an 'or'.)
So let's choose y to be -3.
-3 > -2 FALSE
x + y < 4
x + -3 < 4
x < 7
When x is < 7 it's true. To make it false, we need something bigger than 7. Let's use x = 10
10 + -3 < 4
7 < 4 FALSE
-3 > -2 FALSE
Thus, (10, -3) is not a solution because it makes both inequalities false.
