WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In today's lesson, we are going to be graphing inequalities.
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Recall that, in order to graph a linear inequality, you have to take a series of steps.
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And the graph of a linear inequality is a shaded region in the coordinate plane.
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And those shaded regions are known as half-planes; the boundary of the region, or the half-plane, is a straight line,
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which is the graph of the corresponding linear function.
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The region, or the half-plane, containing the solution set can be determined by using a test point.
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OK, so the steps in order to graph a linear inequality are to first graph the corresponding linear function; and that is a straight line.
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And that line is dashed if the inequality you are working with is a strict inequality (less than or greater than).
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The line is solid if you are dealing with less than or equal to, or greater than or equal to; and we will talk in a minute about what that means.
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After you have graphed the line, then you end up with the coordinate plane divided into two half-planes.
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You use a test point to determine the region containing the solution set.
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So, I am going to go ahead and do an example now: but if you need more review on this, check out the Algebra I video on this topic.
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All right, the easiest way to understand is to look at an example: let's look at the inequality y - 3x < -2.
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My first step is to graph the corresponding linear function, which would be y - 3x = -2.
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So, I am going to go ahead and find some x-values and some y-values--some coordinate pairs on this line.
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When x is 0, y is -2; so that gives me the y-intercept.
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When x is 1, this is going to give me -3; add 3 to both sides--y will be 1.
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When x is 2, -3 times 2 is -6; add that to both sides, and y is 4.
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All right, I have enough to graph the line: x is 0, y is -2--the y-intercept; then I have another point at (1,1) and another point at (2,4).
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Because this is a strict inequality, I am going to use a dashed line.
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This line divides the coordinate plane into two half-planes: an upper and lower half-plane.
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One of these half-planes contains the solution set for the inequality; the way I figure out which one is to use a test point.
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Pick a point that is not on the boundary line, and that is easy to work with.
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(0,0) is very easy to work with; so I am going to take my test point, and I am going to go back to my inequality and plug it in.
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I am going to see what happens: this gives me 0 - 0 < -2; 0 is less than -2--that is not true.
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What this means is that this is not part of the solution set.
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If this is not part of the solution set, it is in the half-plan that is not contained in the solution set.
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Therefore, the solution set is actually in the lower half-plane.
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And you could verify that by checking a point down here.
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The idea is to graph the corresponding linear function to find the boundary line.
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Now, I mentioned that this is a strict inequality, so I am using a dashed line.
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And what that means is that the boundary line is not part of the solution set.
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If it were a solid line (because I had less than or equal to), then the boundary line would be part of the solution set.
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But actually, in this, the solution set is just below that boundary line.
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I determined which half-plane was correct by using a test point.
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Since the test point ended up giving me an inequality that is not true, that is not valid,
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I knew that that test point did not lie in the half-plane of the solution set, so I shaded in the lower half-plane.
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Graphing absolute value inequalities: the procedure for these inequalities is similar to the procedure for linear inequalities.
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So, we are going to use the same technique.
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Recall that the technique involved first graphing the line for the corresponding equation.
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Graph the corresponding equation: in this case, it is going to be an absolute value equation.
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And then, use a test point to determine the half-plane containing the solution set for the inequality.
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All right, let's take an example: y ≤ |x-3|.
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The corresponding equation would be y = |x-3|, so I need to graph that to find my boundary.
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So, some values for x and y: when x is -2, -2 minus 3 is -5; the absolute value is 5.
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When x is -1, this becomes -4; the absolute value is 4; when x is 0, this becomes -3; the absolute value is 3.
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Let's jump up to 3: 3 minus 3 is 0--the absolute value is 0.
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4 minus 3 is 1; the absolute value is 1; 5 minus 3 is 2--the absolute value is 2.
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OK, plotting these points: first, (-2,5), (-1,4), (0,3), (3,0), (4,1), (5,2); OK, I have my typical v-shaped graph of absolute value.
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And you notice that I made this a solid line; and the reason that is a solid line is because I had less than or equal to.
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So, this boundary is actually going to be part of the solution set.
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OK, so I graphed the corresponding equation, and I determined that it is a solid line.
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Now, I need to determine if the solution set is down here, or if it is above the boundary; is it above or below the boundary line?
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So, I am going to use a test point of (0,0): that is going to be my test point.
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I go back to the inequality, and I see what happens when I put my test point values in there.
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This gives me 0 ≤ |-3|; that would say that 0 is less than or equal to 3; that is true.
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Since that is true, what that tells me is that this is part of the solution set.
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(0,0) is part of the solution set, and that means that this lower half-plane where (0,0) is, my test point, contains the solution set.
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So, as I mentioned, this is very similar to the technique for graphing linear inequalities.
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First, graph the corresponding equation; and use a solid line if it is less than or greater than or equal to.
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Use a dashed line if it is a strict inequality (just greater than or less than).
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So then, once you have your boundary line, find a test point that is off the boundary line, and not so close to it as to cause any confusion.
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And choose one that is simple: (0,0), (1,1)--very easy to work with.
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Plug that into the absolute value inequality; and if you come out with a true inequality,
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like this one, you know that you are in the correct area with the solution set.
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If you come out with a statement or inequality that is not valid, you shade in the other region.
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OK, the first example: we have an inequality that we are asked to graph: y ≤ 2x - 3.
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My first step is going to be to graph the corresponding linear equation, in order to determine the boundary.
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Some values for x and y: when x is 0, y is -3; when x is 1, 1 times 2 is 2, minus 3 is -1.
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When x is 2, 2 times 2 is 4, minus 3 is 1.
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OK, so plotting this out: (0,-3), (1,-1), (2,1); I have plotted this boundary line, and I look up here,
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and I see that it is less than or equal to; so I am going to make this a solid line.
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It is going to be a solid line: this line will be part of the solution set.
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So, first I graph the boundary; next, I am going to look at a test point.
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OK, I can again use this very simple test point of (0,0), the origin.
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Going back to the inequality to look at my test point, y ≤ 2x -3; the test point is (0,0), so y is 0; x is 0.
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This is saying that 0 is less than or equal to -3; that is not true--this is not in the solution set.
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Well, the origin is up here; this means that this whole area is not part of the solution set.
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So, I am going to go to the lower half-plane and shade that in to indicate that this shaded region, including the boundary line, is the solution set.
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Again, the first step is to take the corresponding linear equation and graph it.
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The second step: find a test point that is easy to work with and away from the boundary line, and plug that in.
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If you come up with an inequality that is not true, like this, shade the other half.
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If it is true, shade the half containing the test point.
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Example 2: We are graphing another inequality, 2x + y > 6.
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First, I am going to find the corresponding equation: 2x + y = 6.
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I am going to keep this simple: I am going to use the intercept method in order to graph this.
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I am going to let x equal 0 to find the y-intercept: when x is 0, y is 6.
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Then, I am going to let y equal 0 to find the x-intercept: when y is 0, then I am going to end up with 2x + 0 = 6, or 2x = 6; x = 3.
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So, I have my x and y intercepts; it is going to give me...when x is 0, y is 6, right about there; when x is 3, y is 0.
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OK, then I look, and I have a strict inequality, so I am going to have a dashed line;
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this line is going to be dashed, and is not going to be part of the solution set.
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So, I found my boundary line; and that is my first step; my second step is going to be to work with my test point.
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The test point is right here: it is the origin, (0,0).
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Going back and looking at the inequality: 2 times 0, plus 0, is greater than 6; 0 plus 0 is greater than 6.
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Well, 0 is not greater than 6, so this is not true--not in the solution set.
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The origin is not part of the solution set, so the lower half-plane is not the correct half-plane; I am going to shade, instead, this upper half-plane.
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Again, graph the linear equation corresponding to the inequality.
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Since this is strict inequality, plot that line using a dashed line, since this boundary is not part of the solution set.
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Find a test point; the origin is a really good one, as long as it is not on the boundary line or very close to the boundary line.
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Substitute (0,0) into the inequality; and you come up with an inequality that is not true.
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Example 3: Graph 3x - 4 > y.
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First, graph the boundary line: the corresponding equation is 3x - 4 = y.
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To graph that, I am going to need some x and y values; when x is 0, y is -4; when x is 1, that gives me 3 - 4 = y, so y equals -1.
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One more value: when x is 2, y is 2; OK.
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So, I have enough information to plot my line: when x is 0, y is -4; (1,-1), (2,2).
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The next issue: strict inequality--that means this is a dashed line.
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OK, so now, I have this boundary; I have an upper and a lower half-plane; and I need to figure out where my solution set is.
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The origin is well away from the boundary line, so I can use that as my test point.
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So, test point is (0,0): the inequality is 3x - 4 > y; and this gives me 0 - 4 > 0, which comes out to -4 > 0.
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And that is, of course, not true; this is not in the solution set, because this inequality did not hold true.
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Therefore, I am going to go to the other half-plane, and I am going to shade that in.
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I graphed the boundary line using a dashed line, since it is a strict inequality.
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And then, I took a test point, substituted those values into the inequality, and determined that the origin (0,0) is not part of the solution set.
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So, my other half-plane contains the solution set.
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Example 4: we are working with an absolute value inequality where y is less than the absolute value of x - 2, plus 1.
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I am starting out with just thinking that I expect this to look like a v, because absolute value graphs are v-shaped.
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I am graphing the corresponding equation, y = |x - 2| + 1.
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And with absolute value inequalities or equations, I like to get more values,
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because I want to make sure that I don't end up missing half of my v.
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So, when x is -2, that is going to give me -4; the absolute value of that is 4; plus one is 5.
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-1 minus 2 is -3; the absolute value is 3, plus one is 4.
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0 minus 2 is -2; the absolute value is 2, plus one is 3.
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Now, some positive numbers: 1 minus 2 is -1; the absolute value is 1; plus one is 2.
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2 minus 2 is 0, plus one is 1; 3 minus 2 is 1; the absolute value is 1, plus one is 2.
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4 minus 2 is 2; the absolute value is 2, plus one is 3; and now I have enough values to work with.
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So, (-2,5), (-1,4), (0,3), (1,2), (2,1), (3,2)...so you see the v forming, so I know I have enough points,
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because I have both parts of my v...(4,3).
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It's a strict inequality, so I am using a dashed line.
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And you see that it is the typical v shape, but it is shifted over.
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And I have a -2 here, and so this is actually shifted over to the right by 2; and plus 1, so it is shifted up by 1.
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I found the boundary; again, I am fortunate--I can easily use (0,0) as the test point--the origin.
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Going back to the inequality: this is going to be 0 < |0 - 2| + 1.
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So, that is 0 < |-2| (the absolute value of -2 is 2) + 1, so this gives me 0 < 3.
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0 actually is less than 3; therefore, the origin is part of the solution set, so this is a true statement, and the origin is part of the solution set.
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(0,0) is part of the solution set, and that means that I am in the correct half-plane, and I am going to go ahead and shade that in.
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OK, so we handled this just with a very similar technique to linear inequalities.
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Graphing out the absolute value equation, I got my v-shaped graph with a dashed line for the strict inequality.
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I used a test point, (0,0), substituted that into the inequality, and found that I had a true statement--a true inequality, 0 < 3.
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Since that is true, I knew that the origin lay within the region that contains the solution set.
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So, I went ahead and shaded that region to note that my solution set is located there.
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That concludes this lesson of Educator.com; see you next lesson!