Graph the system of inequalities.

y ≥ 2x + 3

y > -x – 3

Begin graphing sequence one on y ≥ 2x + 3. We ignore the inequality sign to find out that the slope is m = 2 and the y-intercept is (0, 3). Then we stop ignoring the sign to see that we have a solid line, because it is a "greater than or equal to" sign.

Let's keep going and find our test point and shading before we graph the line. Our old buddy (0, 0) once again rises to the occasion.

Is 0 ≥ (2)(0) + 3? No, so we shade the other side of our line.

Next comes the graph of y > -x – 3. The slope is -1 and b = -3. We need a dotted line on this one, though.

Testing (0, 0) shows us that 0 > -(0) – 3, so shade above the line. Bring it all together and we get:

The shaded areas that match for both satisfy the system of inequalities.

Graph the system of inequalities.

y ≥ 2x – 1

First we graph .We have a slope of  and b = 2. Our inequality sign gives us a solid line.

Once more we test (0, 0), asking it for guidance. Do we shade above, or below? It doesn't tell us what color to use, though.

Is ? Not at all. Thank you, (0, 0), for your wisdom.

Now y ≥ 2x – 1 is up for graphing. It's another solid line, with a slope of 2 and y-intercept of -1.

Our test point asks us to consider: is 0 > (2)(0) – 1? It is, so we shade this side.

We now combine our graphs to see where the shading overlaps.

These shaded regions are our solutions to the system of inequalities.

Graph the system of inequalities.

y < 4x – 5

Graph number one, , goes 7 below 0 on the y-axis (rather nippy, isn't it?), and rises by 2 and runs by 5. We have a "≤," so we have a solid line.

Using our test point, (0, 0), we check if . It isn't. Shade the other side of our line.

Next, graph y < 4x – 5.

Its y-intercept is -5, its slope is 4, its line is dotted, and its purse is Gucci. Well lah-dee-da, look at how fancy it is.

We test the equation to see which side should be shaded. We're tired of working with (0, 0), so we're going to use (0, 1) instead. Is 1 < 4(0) – 5? No it isn't. We shade the other side.

And now to match up our graphs.

Graph the system of inequalities.

y < -3x + 1

y – 3x < 0

Another day, another set of inequalities to graph. Our first contestant, y < -3x + 1, has a slope of -3, a y-intercept of 1, and a dotted line.

And (0, 0), the wonder point, asks us if 0 < -3(0) + 1. Yes it is, so we shade to the left of the line.

Now y – 3x < 0 is a dotted line graph, with both intercepts at (0, 0). The origin isn't a good test point this time since it's right there on the dotted line, so let's use (1, 0). Is 0 – 3(1) < 0? You bet your britches it is, so we shade the side that includes (1, 0).

Don't these two graphs make a cute couple?

No, they don't? We have to agree to disagree then.