Algebra 1 solve each inequality and graph its solution

To graph a linear inequality in two variables (say, x and y ), first get y alone on one side. Then consider the related equation obtained by changing the inequality sign to an equality sign. The graph of this equation is a line.

If the inequality is strict ( < or > ), graph a dashed line. If the inequality is not strict ( ≤ or ≥ ), graph a solid line.

Finally, pick one point that is not on either line ( ( 0 , 0 ) is usually the easiest) and decide whether these coordinates satisfy the inequality or not. If they do, shade the half-plane containing that point. If they don't, shade the other half-plane.

Graph each of the inequalities in the system in a similar way. The solution of the system of inequalities is the intersection region of all the solutions in the system.

Example 1:

Solve the system of inequalities by graphing:

y ≤ x − 2 y > − 3 x + 5

First, graph the inequality y ≤ x − 2 . The related equation is y = x − 2 .

Since the inequality is ≤ , not a strict one, the border line is solid.

Graph the straight line.

Consider a point that is not on the line - say, ( 0 , 0 ) - and substitute in the inequality y ≤ x − 2 .

0 ≤ 0 − 2 0 ≤ − 2

This is false. So, the solution does not contain the point ( 0 , 0 ) . Shade the lower half of the line.

Similarly, draw a dashed line for the related equation of the second inequality y > − 3 x + 5 which has a strict inequality. The point ( 0 , 0 ) does not satisfy the inequality, so shade the half that does not contain the point ( 0 , 0 ) .

The solution of the system of inequalities is the intersection region of the solutions of the two inequalities.

Example 2:

Solve the system of inequalities by graphing:

2 x + 3 y ≥ 12 8 x − 4 y > 1 x < 4

Rewrite the first two inequalities with y alone on one side.

3 y ≥ − 2 x + 12 y ≥ − 2 3 x + 4 − 4 y > − 8 x + 1 y < 2 x − 1 4

Now, graph the inequality y ≥ − 2 3 x + 4 . The related equation is y = − 2 3 x + 4 .

Since the inequality is ≥ , not a strict one, the border line is solid.

Graph the straight line.

Consider a point that is not on the line - say, ( 0 , 0 ) - and substitute in the inequality.

0 ≥ − 2 3 ( 0 ) + 4 0 ≥ 4

This is false. So, the solution does not contain the point ( 0 , 0 ) . Shade upper half of the line.

Similarly, draw a dashed line of related equation of the second inequality y < 2 x − 1 4 which has a strict inequality. The point ( 0 , 0 ) does not satisfy the inequality, so shade the half that does not contain the point ( 0 , 0 ) .

Draw a dashed vertical line x = 4 which is the related equation of the third inequality.

Here point ( 0 , 0 ) satisfies the inequality, so shade the half that contains the point.

The solution of the system of inequalities is the intersection region of the solutions of the three inequalities.

Video transcript

We're asked to determine the solution set of this system, and we actually have three inequalities right here. A good place to start is just to graph the solution sets for each of these inequalities and then see where they overlap. And that's the region of the x, y coordinate plane that will satisfy all of them. So let's first graph y is equal to 2x plus 1, and that includes this line, and then it's all the points greater than that as well. So the y-intercept right here is 1. If x is 0, y is 1, and the slope is 2. If we move forward in the x-direction 1, we move up 2. If we move forward 2, we'll move up 4, just like that. So this graph is going to look something like this. Let me graph a couple more points here just so that I make sure that I'm drawing it reasonably accurately. So it would look something like this. That's the graph of y is equal to 2x plus 1. Now, for y is greater than or equal, or if it's equal or greater than, so we have to put all the region above this. For any x, 2x plus 1 will be right on the line, but all the y's greater than that are also valid. So the solution set of that first equation is all of this area up here, all of the area above the line, including the line, because it's greater than or equal to. So that's the first inequality right there. Now let's do the second inequality. The second inequality is y is less than 2x minus 5. So if we were to graph 2x minus 5, and something already might jump out at you that these two are parallel to each other. They have the same slope. So 2x minus 5, the y-intercept is negative 5. x is 0, y is negative 1, negative 2, negative 3, negative 4, negative 5. Slope is 2 again. And this is only less than, strictly less than, so we're not going to actually include the line. The slope is 2, so it will look something like that. It has the exact same slope as this other line. So I could draw a bit of a dotted line here if you like, and we're not going to include the dotted line because we're strictly less than. So the solution set for this second inequality is going to be all of the area below the line. For any x, this is 2x minus 5, and we care about the y's that are less than that. So let me shade that in. So before we even get to this last inequality, in order for there to be something that satisfies both of these inequalities, it has to be in both of their solution sets. But as you can see, their solutions sets are completely non-overlapping. There's no point on the x, y plane that is in both of these solution sets. They're separated by this kind of no-man's land between these two parallel lines. So there is actually no solution set. It's actually the null set. There's the empty set. Maybe we could put an empty set like that, two brackets with nothing in it. There's no solution set or the solution set of the system is empty. We could do the x is greater than 1. This is x is equal to 1, so we put a dotted line there because we don't want include that. So it would be all of this stuff. But once again, there's nothing that satisfies all three of these. This area right here satisfies the bottom two. This area up here satisfies the last one and the first one. But there's nothing that satisfies both these top two. Empty set.

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