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In this video, we're going to discuss compound inequalities. Compound inequalities
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are two inequalities that are joined by the word "and" or "or." A value is a solution of a compound
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inequality containing "and" if it is a solution of both inequalities. Let's consider the inequality x
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is less than or equal to five and x is greater than or equal to three. To find the solution set
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of this compound inequality, let's first consider the inequality x is less than or equal to five.
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This would contain five and those points to the left of five on the number line. For the
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inequality x is greater than or equal to three, that solution set would include the value three
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and all of the numbers to the right of three on the number line.
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The solution set to the compound inequality x is less than or equal to five and x is greater
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than or equal to three is the intersection of these two solution sets, meaning that the
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solution set to the compound inequality must include the values that are in both
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solution sets. So that would include the values from three to five, including three and five.
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The places where these solution sets overlap is the solution set to this compound inequality.
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We can also write this compound inequality in the following way: three is less than or equal to x is
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less than or equal to five. Now let's look at some examples of compound inequalities.
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We will solve the compound inequality and graph the solution set on the number line for each
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of these examples. Our first example is negative two is less than two x minus eight is less than six.
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We'll begin by using the Addition Property of Inequalities and adding eight to all three
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sides of this inequality. When I add a number or subtract a number to an inequality, it does not
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change the direction of the inequality sign. So we have six is less than two x is less than fourteen.
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Next, we'll use the Multiplication Property of Inequalities, and we'll divide all three sides by
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two. This is a positive value two, so it does not change the direction of our inequality symbols.
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Dividing all three pieces, or sides, by two gives us three is less than x is less than
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seven. Now, let's graph the solution set on our number line.
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Since my end points here of this inequality contain three and seven those are the values
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that I'll include on my number line, three and seven. Notice that there are no equal to components
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on these inequalities. They are simply less than signs. Therefore, three is not a solution of this
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compound inequality, nor is seven a solution of this compound inequality. So we'll put open circles at
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three and seven, and the solution set of this inequality contains all of the values that are both
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less than seven and greater than three which are all the values in between three and seven.
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In our next example, we have negative one is less than or equal to two-thirds x plus three is
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less than four. Let's begin by subtracting three from all three sides of this linear inequality.
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Subtracting three does not change the direction of our inequality symbols. Negative one
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minus three is negative four. The symbol stays the same less than or equal to
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two-thirds x. And again, the symbol stays the same less than four minus three, which is one.
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And now, we'll multiply by the reciprocal of two-thirds which is three-halves to all three sides.
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Since three-halves is a positive number, this does not change the direction of any of the inequality symbols.
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Multiplying by three-halves here gives us three times negative two, which is negative six.
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Multiplying by three-halves here gives us x. And multiplying by three-halves here gives us three-halves.
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Again, notice that the inequality symbols did not change directions. Now let's graph the
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solution set on a number line. We have negative six and a positive three-halves, which is one and a half.
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Those are the endpoints of this solution set, so I'll put those numbers on my number line.
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Negative six is part of the solution set because of this inequality that has an equal to component.
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Negative six is less than or equal to x, so we'll put a closed circle at negative six. And we'll
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have an open circle at positive three-halves since three-halves is not part of the solution
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set. And the values for x which are both less than three-halves and greater than or equal to
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negative six include the values between those two circles. This is the solution set of the inequality.
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Our third example is zero is less than or equal to four minus x divided by five is less than or
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equal to two. Let's begin by multiplying all three pieces of this inequality by positive
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five. Multiplying by positive five does not change the direction of the sign of these inequalities.
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Zero times five is zero. Multiplying by five on the inside portion of this inequality
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leaves us with four minus x. And multiplying by five on the right side gives us ten. Now we'll use
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the Addition Property of Inequality to subtract four from all three sides of this inequality.
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Zero minus four is negative four. Four minus x minus four is negative x. And ten minus four
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is six. Now I have a negative x in the center of this compound inequality. So to undo this negative
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in front of the x, I'm going to either multiply or divide all three pieces by negative one.
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When you multiply or divide by a negative in an inequality, it does change the direction of these
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symbols. Multiplying negative four by negative one gives us positive four, and we'll flip the sign
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to a greater than or equal to symbol. Negative x times negative one is positive x. We'll flip
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this sign to a greater than or equal to symbol. And six times negative one is negative six.
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Now while this is a true inequality or a correct inequality equivalent to the given statement,
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the standard format here would be to put the smallest number
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on the left side of the inequality. So I will rewrite this inequality as negative six is less
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than or equal to x is less than or equal to four. And now let's graph the solution set.
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The endpoints of the solution set are negative six and four. And because there's an equal to
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component on both of these inequality symbols, we'll put a closed circle at each of those points.
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And the values of x that are both less than or equal to four
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and greater than or equal to negative six are the values here between these two closed circles.
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This is the graph of the solution set of this inequality. Now let's talk about another type of
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compound inequality. A value is a solution of a compound
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inequality containing "or" if it is a solution of either inequality. Let's consider the example
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x is less than or equal to one or x is greater than or equal to three.
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First, we'll begin by graphing the solution set to x is less than or equal to one. So on our number
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line we'll have the numbers one two and three. And we'll begin by graphing x is less than or
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equal to one by putting a closed circle at one and shading those values to the left of one
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because they are less than one. So this is the solution set to x is less than or equal to
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one. Now let's graph the solution set to x is greater than or equal to three.
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We'll begin with a closed circle at three because three is included in the solution set due to the
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equal to component of that inequality symbol. And we'll shade those numbers to the right of three
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on our number line since those are greater than three. So this is the graph of the solution set
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of x is greater than or equal to three. For the inequality that contains x is less than or equal
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to one or x is greater than or equal to three, this graph is the union of these two solution sets,
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meaning the graph should contain all of the solutions from each of the two solution sets.
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So it will contain all of the values here less than or equal to one as well as the values here
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greater than or equal to three. This is the graph of the compound inequality
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x is less than or equal to one or x is greater than or equal to three.
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Let's take a look at these next two examples. We're going to solve the compound inequality for
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each and then graph the solution set on a number line. Our first compound inequality is five x
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minus three is less than or equal to ten or x plus one is greater than or equal to five.
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Recall that a compound inequality has two inequalities combined by an "and" or an "or"
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statement. We'll begin by solving this first piece of this compound inequality. Using the Addition
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Property of Inequality, we'll add three to both sides, giving us five x is less than or equal to
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thirteen. And now we'll use the Multiplication Property of Inequality and divide both sides by
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five. Since five is a positive number, that does not change the direction of the inequality sign.
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This gives us x is less than or equal to thirteen-fifths. Our compound inequality
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contained the word "or." We will continue to write that throughout this process of solving.
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To solve this piece, we'll subtract one from both sides, giving us x is greater than or equal to
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four. So our complete solution here to this compound inequality will be x is less than or
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equal to thirteen-fifths or x is greater than or equal to four. Now let's graph this solution set.
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Thirteen-fifths is two and three-fifths, so this will be between the numbers two and three.
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And I also have the number four that's relevant in this solution set as well. So
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two and three-fifths will be close to the center here. This will be the value thirteen-fifths.
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We'll have a closed circle at thirteen-fifths, and we'll shade all the values that are less than
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thirteen-fifths or to the left of that number on the number line.
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In addition, we'll have all of the values that are greater than or equal to four.
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So we'll have a closed circle at four, and we'll shade all of the values to the right of four.
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Recall that the solution set to a compound inequality containing "or"
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must be a solution to either of these two or the union of these two solution sets.
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Now let's take a look at this compound inequality. Five times x minus two is less than five
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or x plus nine is greater than ten. We'll begin by solving this left piece of the inequality.
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I'll divide both sides by five, which is a positive number thus not changing
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the direction of our inequality sign. And leaving us with x minus two is less than one.
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Then adding two to both sides gives us x is less than three. We'll continue to write "or"
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throughout our solution process. And solving this piece of the inequality, we'll subtract nine from
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both sides, giving us x is greater than one. So the solution to this inequality will contain
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x is less than three or x is greater than one. Now let's graph the solution set on a number line.
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I'll put the numbers one, two, and three on this number line. Let's start
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with x is less than three. That would have an open circle at three and
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all of the values left of three, or smaller than three, would be part of the solution set.
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In addition, we have to include the solution set for x is greater than one. So at one we would have
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an open circle, but notice that at x equals one we already have this as part of the solution set.
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And now we will shade all the values to the right of one which are greater than one. So
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that's going to include this number three and the numbers all the way to the right
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on this number line. Notice we have shaded the entire number line. So every number on this number
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line is a solution to this compound inequality, meaning that the solution set is all real numbers.