I am a little confused with a problem in Introduction to Algebra (from Art of Problem Solving). In the solution it states that for the graph shown in (c), there is an open circle because this graph shows the plane on the graph that satisfies either inequality, but not both and so the point at which they join is excluded. I don't understand why this open circle is used in graph (b) though. This graph shows the planes that satisfy either, or both, inequalities. Why is this point excluded?
To be a bit /more specific:
The two inequalities are:
$2x-4<y$
$y<-\dfrac{2}{3}x+2$
If you graph the region that satisfies either or both inequalities you will end up with a graph as seen in the "Second Image". The point at which both lines meet $(\dfrac{9}{4}, 0.5)$ in the solution has an open-circle indicating that this point is excluded from the region. I am not sure why this point is excluded. Doesn't it satisfy both of the inequalities?
Thanks in advance :)
The two inequalities are $y > 2x - 4$ and $y < - \frac{2}{3} x + 2$.
The lines $y = 2x - 4$ and $y = - \frac{2}{3} x + 2$ intersect at $\left( \frac{9}{4}, \frac{1}{2} \right)$. However, this point does not actually satisfy either inequality. (Remember, the lines for the individual inequalities were dashed, since the points on those lines didn't satisfy the inequalities).
If you plug $(x,y) = \left( \frac{9}{4}, \frac{1}{2} \right)$ into the inequalities you get $\frac{1}{2} > \frac{1}{2}$ and $\frac{1}{2} < \frac{1}{2}$. Neither of those is true! Since that point doesn't satisfy either of your inequalities, it is left as an open circle on the graph.