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 discussing solving inequalities.
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Recall the properties of inequalities: these are needed to solve inequalities.
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The first one is that, if a is greater than b, and you add the same number to both sides, this inequality has the same solution set as the original inequality.
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And you can think of this using numbers to make it clearer.
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If I said that 6 is greater than 5, and then I decided to add 2 to both sides, 8 is greater than 7; so this still holds up.
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You are allowed to add the same number to both sides of an inequality; and that is true for both greater than and less than.
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And it is also true for greater than or equal to and less than or equal to.
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The subtraction property states that, if a is greater than b, and I subtract the same number from both sides,
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then the resulting inequality has the same solution set as the original inequality.
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The same idea as up here: if 6 is greater than 5, and I subtract 3 from both sides, I am going to end up with 3 > 2; that still holds up.
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And that also is valid for less than, and for greater than or equal to and less than or equal to.
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With multiplication properties, if a is greater than b, and a positive number (c > 0), the same positive number,
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is multiplied by both sides of the inequality, then the solution set of the resulting inequality is the same as that of the original inequality.
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If I have 3 > 2, and I multiply both sides by 4, I am going to end up with 12 > 8, which is still valid.
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The same for less than: a < b--you are allowed to multiply both sides of the inequality by the same positive number, without changing the solution set.
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Again, this applies to greater than or equal to and less than or equal to, as well as to strict inequalities.
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The case is different for negative numbers: if c is less than 0 (if you were multiplying both sides of an inequality
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by a negative number), you must reverse the inequality sign.
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If both sides of an inequality are multiplied by the same negative number, the direction of the inequality must be reversed.
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And if you do that, then the resulting inequality has the same solution set as the original inequality.
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If you don't reverse them, the solution set will not necessarily be the same.
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For example, if I have 8 > 4, and I multiply both sides by -2, let's say I didn't reverse the sign:
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then, I am going to end up with -16 > -8; and this is clearly not true.
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So, as soon as I multiply by a negative number, this is incorrect; I immediately have to reverse the direction of the inequality.
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And that will give me -16 < -8, which is valid; and again, the same for greater than or less than or equal to.
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So, multiplying by a positive number, the solution set of the resulting inequality is the same as the original.
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Multiplying by a negative number, you need to reverse the inequality sign in order for the solution set for this inequality to be the same as for the original.
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A similar idea with division: if you are dividing both sides of an inequality by a positive number,
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the resulting inequality has the same solution set as the original.
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That is only if it is for a positive number; so 15 > 20--if I wanted to divide both sides by 5, I could do that.
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And I don't need to do anything to the inequality; I just keep it the same.
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And that is going to give me 3 > ...well, I need to start out with an inequality that is valid!
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So, if 20 > 15, and then I divide the same number by both sides...so I am starting out with 20 > 15;
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20 divided by 5 is greater than 15 divided by 5, which is going to give me 4 > 3, which is true.
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The same holds up for less than, and greater than or equal to/less than or equal to.
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Now, if we are dealing with dividing by a negative number, you have to reverse the inequality sign.
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And then, that resulting inequality is the same solution set as the original inequality.
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So, if I have 4 < 6, and I divide both sides by -2, I need to immediately reverse this inequality symbol.
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And then, I am going to get 6 divided by -2, and I am going to end up with -2 > -3; and that is valid.
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If I hadn't reversed this, I would have gotten something that is not valid, or does not have the same solution set as the original.
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So, you need to be very careful, as soon as you multiply or divide by a negative number,
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when you are working with inequalities, that you reverse the inequality symbol.
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There are multiple ways of describing the solution set of an inequality.
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You can either express it as a graph, a set, or simply as an inequality.
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For example, if I came up with the solution set x ≥ 2, I could just leave it as an inequality like that; that is a little less formal.
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I could describe it as a set, using set builder notation.
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And I would put it as follows: what this is saying is "the set of all x, such that x is greater than or equal to 2."
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So, this is set builder notation: the set of all x, such that x is greater than or equal to 2.
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You will see the same things, sometimes, with two dots here; this is more common, but you will see this sometimes; and it means the same thing.
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You can also use a graph on the number line; you can express it as a graph.
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And remember that, if you are saying "greater than or equal to," you are including this number 2 in the solution set.
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So, in that case, you would want to use a closed circle, and then continue on to the right.
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Now, let's say I was going to say x is less than 3.
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In this case, it is a strict inequality; and 3 is not part of the solution set, so I am going to use an open circle.
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So, you can express an inequality's solution set either using set builder notation, using a graph, or less formally, just as a simple inequality.
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Looking at Example 1, we have -3x < -27, so I need to solve that.
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I need to isolate this x; and in order to do that, I am going to need to divide by -3.
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But as soon as I start thinking about dividing by a negative number, I have to reverse the inequality symbol.
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So, instead of less than, this becomes greater than.
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This is going to give me x > 9.
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I can leave it just as an inequality; I can use set notation to show this; I can also graph it on the number line.
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OK, and graphing it on the number line, I am going to use an open circle (since it is a strict inequality)
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and show that x is greater than 9, but that 9 is not part of the solution set.
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Solve -4x - 7 ≥ 9: the first step is to add 7 to both sides, to get -4x ≥ 16.
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Now, to isolate x, I need to divide both sides by -4; and as soon as I do that, I am going to change this direction to less than or equal to.
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16 divided by -4 is -4, so the solution is x ≤ -4.
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Using set notation, and graphing on the number line--multiple ways...-1, -2, -3, -4.
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And this is saying "less than or equal to"; therefore, -4 is going to be included in the solution set, as indicated by a closed circle.
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OK, in Example 3, it is a little bit more complicated; but you just handle it using the same principles.
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When you are dealing with fractions, the best thing to do is to get rid of them first, because they are difficult to deal with.
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So, I am going to multiply both sides of the equation by -9, and (since that is a negative number)
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immediately reverse the inequality symbol, so that you don't forget to do that.
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This is going to be -9 times -2 times 3x - 6; I'll move this over a bit...and that is +(-9), times 4x, divided by -9.
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It is greater than -3 times -9.
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OK, so -9 times -2 is 18, times 3x - 6...these -9's cancel out, and that is going to give me...+4x, is greater than (-3 times -9 is) 27.
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Using the distributive property, I am going to multiply this out; and 18 times 3x is 54x; 18 times -6 is -108; plus 4x, is greater than 27.
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Now, I need to isolate this x; so first, I am going to combine like terms.
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I have 54x + 4x, is 58x, -108 is greater than 27.
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Adding 108 to both sides will give me 58x > 27 + 108, so that is 58x > 135.
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So, x is greater than 135/58; I can leave it like that, or I can use set notation.
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To graph this, you need to figure out...if you divide 135/58, it is a little bit larger than 2; it is approximately 2.3.
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So, I could go ahead and graph that out, as well.
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2 is here; 2.5 is about there; 2.3 is about right here; open circle at 2.3, or actually right over here;
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1 is here; 2 is here; 2.5 is about here; so, 2.3 is going to be right about here; and that is open circle, like that.
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There are three ways to express the solution.
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OK, in Example 4, again, it is a little bit more complex, and there is a fraction, so we are going to get rid of that first.
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Start out by multiplying both sides by -7, which tells me that I immediately need to reverse the sign.
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-7 times -3 times (x - 4), over -7, plus -7 times 3 times (9 - 2x); reverse the inequality symbol--greater than or equal to -7 times (4 - x).
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OK, this cancels; that gives me -3 times (x - 4), and this is -7 times 3, so that is -21, times (9 - 2x), is greater than or equal to...
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-7 times 4; that is -28; -7 times -x is plus 7x.
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So now, I just need to do some more simplification.
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-3 times x is -3x; -3 times -4 is +12; -21 times 9 is -189; -21 times -2x is +42x.
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And this is -28 + 7x; I can't really do anything with that.
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Now, I am going to combine like terms (and I do have some like terms).
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I have a -3x and 42x; combining those, I am going to get 39x; -189 +12 is -177.
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The next step is to isolate the x, so I am going to add 177 to both sides; that is going to give me 177 - 28 + 7x.
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Subtract 7x from both sides: 39x - 7x ≥ 177 - 28.
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Combine like terms to get (39x - 7x is) 32x ≥ 149.
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Now, I am going to divide both sides by 32; and it is a positive number, so it just becomes x ≥ 149/32.
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Using set notation, x is greater than or equal to 149/32.
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You could also graph this; this is approximately equal to 4.7, so 0, 1, 2, 3, 4, and 5;
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it is going to be a little over halfway between 4 and 5; closed circle; and graph it.
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OK, so the first step was to eliminate the fraction by multiplying both sides of the equation by -7 and reversing the inequality symbol.
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Once that was done, we are using the distributive property to multiply everything out and get rid of the parentheses.
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Then, we are using the addition and subtraction principles and combining like terms to simplify this inequality.
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And the result was x ≥ 149/32.
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That concludes this lesson on solving inequalities at Educator.com; see you again!