# System of Inequalities

## System of Inequalities - How it Works - Video

### Example 1 Example 1:

We have a system of inequalities. It is just like a system of equations but we have to shade. We have to two inequalities x + y < 2 and 2x - y 4. Our first step is to graph the one of the inequalities. Let's start with x + y < 2.

We found two points (2, 0) and (0, 2) to graph our line. Since we do not have or equal to, we have a dotted line. Now we pick any two points, but one must be above the line and the other below the line. We have picked (0, 0) and (2, 2). Once we picked our two points we substitute them into the inequality. One will be true and one will be not true. We shade the entire section that contains the point that makes our inequality true.

In this case the point (0, 0) makes the x + y < 2, true so we shade below the dotted line. We found two points (0,-4) and (2, 0) to graph our second inequality, 2x - y 4. Since we do have or equal to, we have a solid line.

Now we pick any two points, but one must be to the left of the line and the other to the right of the line. We have picked (1, 2) and (4, 2). Once we picked our two points we substitute them into the inequality. One will be true and one will be not true. We shade the entire section that contains the point that makes our inequality true.

In this case the point (1, 2) makes the 2x - y 4, true so we shade to the left of the solid line.

Right now we have two sections that are blue and two sections that are yellow. One of the yellow sections in behind the blue section. What does this mean? What you see above is the final answer. Since we have 2 inequalities, we need both to be true at the same time. In order for that to happen, we leave the section that is shared by both.

One blue section overlapped one yellow section; blue and yellow make green so we leave that section to show where it is always true for both.

To verify that is true you should always pick a point in the section and substitute it into both inequalities. (0, 0) is the usually the best point and it is for this case too. Since 0 < 2 and 0 4, we shaded the correct section.

### Example 2 Example 2:

We have a system of inequalities. It is just like a system of equations but we have to shade. Woe! We have 4 inequalities this time: x + y < 5, 2x - y < 4, x >1, and y >1. Just like in example 1 our first step is to graph the one of the inequalities. Let's start with x + y < 5.

We found two points (0, 5) and (5, 0) to graph our line. Since we do not have or equal to, we have a dotted line. Now we pick any two points, but one must be to the left the line and the other to the right of the line. We have picked (1, 1) and (4, 3). Once we picked our two points we substitute them into the inequality. One will be true and one will be not true. We shade the entire section that contains the point that makes our inequality true.

In this case the point (1, 1) makes the x + y < 5, true so we shade to the left of the dotted line. We found two points (0, -4) and (2, 0) to graph our second inequality, 2x - y < 4. Since we do not have or equal to, we have a dotted line.

Now we pick any two points, but one must be to the left of the line and the other to the right of the line. We have picked (2, 4) and (4, 2). Once we picked our two points we substitute them into the inequality. One will be true and one will be not true. We shade the entire section that contains the point that makes our inequality true.

In this case the point (2, 4) makes the 2x - y < 4, true so we shade to the left of the dotted the dotted line.

Right now we have two sections that are purple and two sections that are yellow. One of the yellow sections in behind the purple section. What does this mean? What you see above is the final answer. Since we have 2 inequalities, we need both to be true at the same time. In order for that to happen, we leave the section that is shared by both.

One purple section overlapped one yellow section; purple and yellow make dark brown (we went with light brown so we can see the numbers easier) so we leave that section to show where it is always true for both.

To verify that is true you should always pick a point in the section and substitute it into both inequalities. (0, 0) is the usually the best point and it is for this case too. Since 0 < 5 and 0 < 4, we shaded the correct section.  We found two points (1, 2) and (1, -3) to graph our third inequality, x > 1. Since we do not have or equal to, we have a dotted line.

Now we pick any two points, but one must be to the left of the line and the other to the right of the line. We have picked (-2, -3) and (2, -3). Once we picked our two points we substitute them into the inequality. One will be true and one will be not true. We shade the entire section that contains the point that makes our inequality true.

In this case the point (2, -3) makes the x > 1, true so we shade to the right of the dotted the dotted line.

If we count the number of sections that are created with just the dotted lines, we have 7 sections. Right now we have two sections that are light brown, 1 and 4, and four sections that are orange, 3, 4, 5, and 7. Only one section is shared by both light brown and orange.

What does this mean? What you see above is where it is always true for x + y < 5, 2x -y < 4, and x >1. We still have one inequality to graph, y > 1.

One light brown section, 4, overlapped one orange section, 4; light brown and orange make a reddish color so we leave that section to show where it is always true for all three.

To verify that is true you should always pick a point in the section and substitute it into the three inequalities. (0, 0) is the usually the best point, but this time (0, 0) is not in the middle of the shaded section so we have to choose another point.

Let's choose (2, 1). Since 3 < 5, 3 < 4, and 2 > 1, we shaded the correct section.  We found two points (-4, 1) and (-1, 1) to graph our fourth inequality, y > 1. Since we do not have or equal to, we have a dotted line.

Now we pick any two points, but one must be above the line and the other below the line. We have picked (-2, 2) and (-2, -1). Once we picked our two points we substitute them into the inequality. One will be true and one will be not true. We shade the entire section that contains the point that makes our inequality true.

In this case the point (-2, 2) makes the y > 1, true so we shade to the above the dotted line.

If we count the number of sections that are created with just the dotted lines, we have 9 sections. Right now we have two sections that are reddish, 4 and 7, and five sections that are light green, 1, 2, 3, 4, and 5. Only one section is shared by both the reddish one and the light green.

What does this mean? What you see above is where it is always true for x + y < 5, 2x -y < 4, x >1, and y > 1.

One reddish section, 4, overlapped one light green section, 4; so the final section is where all the sections overlapped.

To verify that is true you should always pick a point in the section and substitute it into all four inequalities. (0, 0) is the usually the best point, but this time (0, 0) is not in the middle of the shaded section so we have to choose another point.

Let's choose (2, 2). Since 4 < 5, 2 < 4, 2 > 1, and 2 > 2 we shaded the correct section.