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Solving Systems of Linear Inequalities by Graphing

Solving Systems of Linear Inequalities by Graphing

Author: Colleen Atakpu
Description:

This lesson will demonstrate how to solve systems of linear inequalities by graphing. 

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Today we're going to talk about solving a system of linear inequalities by graphing. So we're going to start by reviewing how to graph a single linear inequality, and then we'll do some examples graphing systems of linear inequalities.

So let's review how to graph a linear inequality. If I were to replace this inequality sign with an equals sign, I would have an equation in slope intercept form. So I'm going to graph this by finding the slope and the y-intercept. So my y-intercept is negative 5. So I'll place a point on my y-axis at negative 5. And because my slope is 1/4, from my y-intercept I'm going to go up 1 and over 4 to find another point. Now I notice that my inequality symbol is less than or equal to, so I know that I'm going to use a solid line to connect my points. Similarly, if my inequality symbol was greater than or equal to, I would use a solid line. But if the inequality symbol was strictly less than or strictly greater than, I would use a dotted line. I then notice that my inequality symbol is less than or equal to, so I know that I'm going to shade below the line. If it were an inequality symbol of strictly less than, I would also shade below my line. But if the inequality symbol had been greater than, or greater than or equal to, I would have shaded above my line.

So let's see how we can solve a system of linear inequalities by graphing. Remember, the solution to a system of inequalities satisfies all of the inequalities in the system. And so by looking at a graph, the solution is going to be the area or region that is overlapped by all of the regions represented by each inequality in the system. So let's, for our first example, look at the system-- y is less than or equal to negative 3, and y is less than 1/3 x minus 2. If I were to replace the inequality symbol with an equals sign, I would have y is equal to negative 3, which I know is a horizontal line through negative 3 on my y-axis.

So I'm going to graph that line, and that line is going to be the boundary of my shaded region for the first inequality. So I'm going to find negative 3 on my y-axis and draw a horizontal line through negative 3. Looking at my inequality symbol, I know that because it's a less than or equal to instead of just strictly less than, I know that I'm going to use a solid line for the boundary of my shaded region. Again looking at the inequality symbol, because it's less than or equal to, instead of greater than or equal to or greater than, I know that I'm going to shade below my boundary line.

All right, let's graph the second inequality. If I were to replace this inequality symbol with an equals sign, I'd have an equation that is written in slope intercept form. So I see the y-intercept would be negative 2 and the slope would be 1/3, so I'm going to start graphing my inequality by graphing that line, which again will be the boundary of my shaded region. So I'll place a point at negative 2 on my y-axis for my y-intercept. And then I'll use a slope of 1/3 from that point. So I'll go up 1 and over 3. Looking at my inequality symbol, it's strictly less than, instead of less than or equal to, so this time I'm going to use a dotted line or dashed line to connect my points. And again because it is less than, instead of greater than or greater than or equal to, I know that I'm going to shade below that line.

So looking at the area-- both the red and the blue region overlap. I'll see my solution is this region here, and I'll shade that in green. So this shaded region in green is the solution to my system of inequalities. I can verify that by selecting a point within that shaded region, and using the x and y values from that point to substitute them into my inequalities and make sure that they satisfy both inequalities. So I'm going to pick the point here, which is the point 0 negative 4, and I'm going to substitute 0 in for x and negative 4 into y into both inequalities to see that those values satisfy both inequalities. So for the first one, I'll substitute negative 4 in for y. And I don't have an x variable. I just have negative 4 is less than or equal to negative 3, which is true. Negative 4 is less than negative 3. Now I'll check my second inequality, substituting negative 4 in for y and 0 in for x. I'll simplify this-- 1/3 times 0 is 0, 0 minus 2 is negative 2, and negative 4 is less than negative 2.

So let's look at a second example. I've got another system of inequalities-- y is greater than x, and y is less than or equal to x minus 3. My first inequality-- y is greater than x-- If I were to replace this inequality symbol with an equals sign, I'd have the equation y is equal to x, which is written in slope intercept form. There's no value for my y-intercept, which means that it's equal to 0, so I know that this will have a y-intercept of 0. And the slope of the line is going to have a slope of 1. If there's no number in front of the x, then we assume it's a 1-- or we know that it's a 1.

So I'm going to start graphing by placing a 0 on my y-axis. And then using my slope of 1, which is the same as 1/1, I'll go up 1 over 1 to get a second point, and 1/3. And now I can graph this line. I know that since I'm actually graphing the inequality, I have to look at the inequality sign. And since it is strictly greater than, instead of greater than or equal to, I'm going to use a dash line to connect my points. And then to determine the region that I need to shade, I can look at my inequality symbol. Since it's greater than, instead of less than or less than or equal to, I'm going to shade above my line.

For my second inequality-- again, if I were to replace this symbol with an equals sign, I'd have another equation in slope intercept form. It would have a y-intercept of negative 3 and a slope of 1, or 1/1. So to graph it, I'll start at negative 3 on my y-axis. Then using my slope of 1/1, I can get a couple more points. Looking at my inequality symbol-- because it's less than or equal to, I know that I'm going to use a solid line to connect my points. And then to determine the shaded region, the region that I should shade, I notice that it's less than or equal to, instead of all greater than or greater than or equal to, which means I'm going to shade below the line. So now I'm looking for again the region that is overlapped by both shaded regions for my system. But there are no regions of overlap because these two lines are parallel-- they're never going to cross. And so the solution to this system of inequalities is no solution, because there is again no region of overlap between these two inequalities.

So let's go over our key points from today. The solution to a system of linear inequalities on a graph is the region of overlap of all shaded regions. In shading regions for vertical lines, the symbols less than and less than or equal to indicate shading to the left of the line. And the symbols greater than and greater than or equal to indicate shading to the right of the line. In shading regions for horizontal lines, the symbols less than and less than or equal to indicate shading below the line. And the symbols greater than and greater than or equal to indicate shading above the line.

So I hope that these key points and examples helped you understand a little bit more about solving systems of linear inequalities by graphing. Keep using your notes, and keep on practicing, and soon you'll be a pro. Thanks for watching.