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 continue on with talking about rational expressions.
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this time extending our discussion to solving rational equations and rational inequalities.
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The definition of a rational equation is an equation that contains one or more rational expressions.
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And remember that rational expressions are algebraic fractions.
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For example, a rational equation could be something like this: (x² - 3x + 9)/(x² - 36) + 3/(x + 5) = 1/2.
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The technique for solving rational equations is going to be to eliminate the fractions.
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And that can be done by multiplying each term of the equation by the least common denominator of all the fractions in the equation.
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We are going to find the least common denominator, and then multiply each term in the equation by that expression.
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Recall, from previous lectures: we discussed the least common denominator.
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And, if necessary, go back and review that, because we are going to be applying that idea to solve rational equations.
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Let's look at another, simpler example, and then go ahead and solve it.
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2 divided by (x + 3), minus 1/2x, equals 1/x.
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Here is a rational equation, and according to this technique, what I need to do is find the least common denominator of all the fractions in the equation.
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Recall that the LCD is the product of all the unique factors in the denominator, to the highest power that they are present in any one polynomial.
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Therefore, if I look at the denominators, they are already factored out; that makes this very simple.
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I have (x + 3), 2x, and x; therefore, the LCD would be (x + 3)...that is a unique factor; in this second denominator,
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I have 2 times x; so 2 and x are both factors; and then I have x--however, x is already represented here.
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And I only need to represent each unique factor the highest number of times that it is present in any one of the denominators.
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So, x is present here once, and it is present here once; so I only have to represent it here once.
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If I had x², for example, here, then I would have written this as 2x²(x + 3).
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Once I have my least common denominator, I am going to go back and follow this technique, multiplying each term of the equation by the LCD.
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And then, we will see what happens.
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For the LCD, 2x times (x + 3), times 2 divided by (x + 3), minus 2x times (x + 3), times 1/2x, equals 1/x times 2x(x + 3).
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Let's start canceling out to simplify: the (x + 3)'s cancel here, leaving me with 2x times 2, minus...
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here, the 2x's cancel out, leaving me with (x + 3) times 1, or just (x + 3), equals...
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here the x's cancel out, leaving 2 times (x + 3).
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At this point, you can see that the fractions have been eliminated, which was exactly my goal.
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So now, I am just going to set about simplifying this.
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2x times 2 is 4x; minus x, minus 3, equals 2x + 6.
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All I have here now is a linear equation, and I just need to solve for x.
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4x - x is 3x; minus 3 equals 2x + 6; I am going to subtract 2x from both sides: that is going to give me x;
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and at the same time, I am going to add 3 to both sides; so moving the variables to the left,
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and the constants to the right, gives me x = 9.
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And I always want to just double-check and make sure that this is not going to cause the denominator to become 0, because that is not allowed.
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And we will talk more about values that would cause the denominator to be 0 in a few minutes.
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But for right now, I can see that 9² - 36 would not give me 0, and 9 + 5 would not give me 0.
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So, this is a valid solution.
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Again, the technique is just to find the LCD and multiply each term in the rational equation by the LCD, in order to eliminate the fraction.
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Now, it actually would have been possible to solve this by converting all of these rational expressions,
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all of these fractions, to a common denominator, and then just adding and subtracting.
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But that is actually a lot more work: it is much easier to go about it this way and just get rid of the fractions.
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Let's do another example to illustrate this technique.
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5/(x² + x - 2) + 4/(x - 1) = 1/(x + 2).
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Recall that the first step is to find the LCD.
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In order to do that, I am going to go up here and factor out each of the denominators.
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x² + x - 2: this is going to factor to (x + something) (x - something).
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And I want factors of 2 that add up to the coefficient of 1 in front of the x.
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Therefore, I am going to put the 2 here and 1 here; and this will give me x² - x + 2x, which will give me x;
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and then, the last term is - 2; so that checks out.
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x - 1 just stays as x - 1; and x + 2 is x + 2.
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So, the LCD: I have this factor, x + 2, and the highest number of times it is present in any one denominator,
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in any one polynomial, is once, so I can leave it as (x + 2).
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(x - 1) is also present only once in each of these two, so I am going to represent it once.
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The LCD is (x + 2) (x - 1).
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I am rewriting this with the denominators in factored form, (x + 2) (x - 1), and leaving some room to multiply by the LCD.
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4/(x - 1) = 1/(x + 2).
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I am going to multiply each term by the LCD: (x + 2) (x - 1).
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Again, here, I am multiplying this by (x + 2) (x - 1); and finally, on the right side, (x + 2) (x - 1).
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The goal is to get rid of the fractions, so let's check out if that worked.
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These cancel; both factors cancel, leaving behind only a 5.
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Right here, the (x - 1)s cancel, leaving behind 4 times (x + 2).
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On the right side, the (x + 2)s cancel, leaving behind 1 times (x - 1), or simply (x - 1).
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Now, I am left with a linear equation that I can solve by isolating x.
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5 + (4x + 8) = x - 1.
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Combining the 8 and the 5 is going to give me 4x + 13 = x - 1.
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I am going to subtract x from both sides to combine the variables on the left, so 4x - x is 3x + 13 = -1.
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Now, I am going to go ahead and subtract 13 from both sides to give me 3x = -14.
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And then, dividing both sides by 3 gives me x = -14/3, which I can either leave like that or convert to a mixed number, -4 2/3.
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So again, the technique to solve a rational equation is to find the least common denominator,
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multiply each term in the equation by the LCD to get rid of the fractions, and then proceed as you would to solve an equation.
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All right, I mentioned a little bit ago that you have to watch out and make sure that the solution that you have is valid.
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Now, we are going to talk about that in detail.
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You might recall that we talked about extraneous solutions when we talked about radical equations
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(equations containing radicals that had a variable as part of the radicand).
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And we used a technique there where we squared both sides of the equation.
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But the result could be the introduction of solutions that were not valid--extraneous solutions.
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Here, the same thing can happen: the technique that we are using will find a solution if one exists.
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But extra solutions can also pop up that are not valid.
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So, using this technique, you might find that you have one extraneous solution and one valid solution;
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all valid solutions; all extraneous solutions; it could be any combination--you just have to check.
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And if you find that none of the solutions are valid, it means that there is no solution,
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because if a solution exists, you will find it with this technique; but it might be buried among extraneous solutions.
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If you find that all of the solutions are extraneous, that means there is no solution to the equation.
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Let's take an example: (x + 1) divided by (x² + 2x + 15)/(x - 3) = 1/5x.
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Let's just focus on possible extraneous solutions without solving out this whole thing.
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To find the extraneous solutions, you need to think about values
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that would make the denominator 0, because a denominator of 0 is undefined; it is not allowed.
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So, if we ended up with a solution that made the denominator 0, that solution would be extraneous; it would be invalid.
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Factoring out this first denominator, x² + 2x - 15, is going to give me (x + something) (x - something).
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And factors of 15 are 1 and 15, 3 and 5; and I am looking for factors that will add up to 2.
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5 and -3 would add up to 2; so I am going to go with 5 here and -3 there.
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Recall that the problem would occur if there was a value of x such that this polynomial became 0.
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So, if I had (x + 5) (x - 3) = 0, a value of x that gave me that would be an extraneous solution.
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And using the zero product property, you can say that, if either (x + 5) is 0 or (x - 3) is 0, this whole expression becomes 0.
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So, I am going to set (x + 5) equal to 0 and (x - 3) equal to 0, and solve.
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So, if x equals -5, or if x equals 3, either way, those would be values that are not allowable; and they would be excluded.
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OK, so these are some excluded values; and these are based on that first denominator.
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Looking at the second denominator: if x - 3 = 0, I would also end up with a 0 in the denominator, and that is not allowed.
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So, solving for x, x = 3; however, I already have that accounted for here, so I don't have to write that again under my list.
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5x--I cannot let 5x equal 0; and the situation where that would equal 0 is if x itself equaled 0, so another excluded value would be x = 0.
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My excluded values are if x equals -5, 3, or 0.
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One way to approach this is to find the excluded values first, when you are working with a rational equation.
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And then, go through and solve, and when you get to your solution, check and make sure it doesn't match an excluded value.
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If it does, you have to throw that solution out.
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If it doesn't match an excluded value, then you are left with a solution that is valid.
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Rational inequalities are a little bit more complex to solve; so we are going to go through, and then we are going to do an example of how to solve these.
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Recall that a rational inequality is an inequality that contains one or more rational expressions.
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We talked about rational equations containing algebraic fractions, or rational expressions.
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The idea is the same here, except we are dealing with inequalities instead of equalities.
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There are several steps to solve these: the first step is to find the excluded values.
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We talked about excluded values when working with rational equations; and it is the same idea here.
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It is going to be any value of x that is going to make the denominator 0.
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Then, the second step is to solve the related equation.
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Your result is going to be that you are going to have some values of x from #1; you are going to have values of x from #2;
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and then, you are going to take the values from steps 1 and 2 and use them to split the number lines into intervals.
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Those intervals may contain parts of the solution set; so then, you are going to use a test value for each interval.
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And that test value is going to help you determine whether or not the interval contains parts of the solution set.
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An example of a rational inequality would be something like x/(x² - 9) - 3/(x + 4) < 5.
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We will do an example in a second; but pretty much, what you are going to do is find the excluded values
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by finding values that would make the denominator 0, then solve the corresponding equation
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by changing this into an equal sign, and then solving, taking those values (whatever those values are), and then dividing the number line.
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Let's say I ended up with two solutions and two excluded values; then I would divide the number line into 1, 2, 3, 4, 5 intervals.
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And then, I would use test points to determine if these intervals contain parts of the solution set.
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I would use the test values and insert them into the original inequality, and see if the inequality held out, based on those values.
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Let's illustrate this right now with an example.
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Given the rational inequality x/(x² - 9)...actually, let's do a slightly less complex one.
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3/x + 1/(x + 4) ≥ 0
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I am going to start out by finding the excluded values--values that would make the denominator 0.
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For this first one, it is very simple: if x equals 0, then this becomes the denominator 0, and that is not allowable.
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The second one: I get x + 4 = 0; if that were true, this would become an undefined expression.
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Solving for x, x equals -4: if x is -4, add that to 4 and you get 0.
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So, I have two excluded values: x = 0 and x = -4; that is my first step.
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My second step is to solve the related equation, which is 3/x + 1/(x + 4) = 0.
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Recall that we talked about solving rational equations by multiplying all terms by the LCD.
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And these are already factored out; so my LCD is the factor in this denominator, which is unique, times the factor in this denominator,
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x times (x + 4)...so I am going to use the technique of multiplying each term by this LCD to get rid of the fractions.
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Right now, all I am doing is trying to solve this, so that I can add it to this set of values, and then split up the number line into intervals.
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That is my first term, x(x + 4) times 1/(x + 4) = 0 times x(x + 4).
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I see here that the x's cancel out, leaving me with 3(x + 4) +...here the (x + 4)s cancel out...that is x =...this whole thing becomes 0.
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Now, I need to solve this linear equation: this gives me 3x + 12 + x = 0.
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3x + x is 4x, plus 12 equals 0; subtract 12 from both sides to get 4x = -12; and then finally, divide both sides by 4 to yield x = -3.
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And that is not an excluded value, so that is a valid solution to this equation.
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So, I have excluded values, and I have a solution to the related equation: x = -3.
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The next step is to use these to divide up the number line into intervals.
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Let's do that down here, out of the way: here x = 0, -1, -2, -3, -4; so, -4: -1, -2, -3...
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Now that I have the number line divided up, I need to determine which of these intervals...
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I have one interval here; I will call this A; it is from right here...this is also dividing it into an interval...
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eliminate these to make the intervals clearer; I have interval B, C, and D; so I have 4 intervals.
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I need to find test points and check those in the original inequality for all 4 of these.
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For the first interval, I am going to use a test point of -5.
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I am working right over here; let's let x equal -5.
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If I go back to this original inequality, let's see if this value over here satisfies it: 3/-5 + 1/(-5 + 4) ≥ 0.
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This gives me -3/5; and this becomes 1/-1, because -5 + 4 is -1.
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So, this gives me -1 3/5, which is not greater than or equal to 0.
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Therefore, this interval does not contain any of the solution set, because my test point did not satisfy this inequality.
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This was for interval A: for interval B, I am going to pick another point, and that is between -3 and -4.
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So, I can't pick an integer; I have to go with a fraction to get something in between there.
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So, I am going to pick -3 1/2; so let's let x equal -3 1/2, which actually equals -7/2.
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This is going to give me 3, divided by -7/2, plus 1, divided by (x + 4), is greater than or equal to 0.
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3 divided by -7/2...I could simplify this complex fraction by turning this into 3 times -2/7; so this becomes (I'll write it over here) 3 times -2/7.
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Actually, this should be -7/2 right there; and -7/2 + 4 (let's work on this right here)...that would give me...
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I need to convert this, so this would give me 8/2, because 8 divided by 2 would give me 4 back.
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So, that is a common denominator of 2; -7/2 + 8/2 = 1/2.
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That is 3 times -2/7, plus 1/(1/2), is greater than or equal to 0.
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This gives me -6/7; getting rid of this complex fraction, 1 divided by 1/2 is equal to 1 times 2, which is 2.
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-6/7 is smaller than 2, so I know that this becomes positive; therefore, this is true.
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When I use x = -3/2, I find that I satisfy the inequality; therefore, B contains part of the solution set.
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So, this value did not satisfy the inequality; this value did satisfy the inequality.
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Looking at...we are going to move over here...the third interval: for the third interval, I am going to use a value of -2,
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because it has to be between 0 and -3; so let's go ahead and use -2.
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When x equals -2, that gives me 3 divided by -2 plus 1, divided by -2 plus 4, is greater than or equal to 0.
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And that becomes -3/2; and this is +1/2, is greater than or equal to 0.
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1/2 and -3/2 gives me -2/2, or -1; is that greater than or equal to 0? No, it is not.
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So, C does not contain the solution set.
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Finally, I am going to test section D, using the test point of 1.
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Let's let x equal 1; therefore, this is going to give me 3 divided by 1, plus 1 divided by (1 + 4), is greater than or equal to 0.
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And I can see that these are all positive numbers, so even if I don't figure this out, I know that this is valid; so yes.
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I have determined that the intervals B and D contain the solution set.
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Now, as far as how to write this solution set: let's go up here and write it out.
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You need to be careful that you don't include excluded values.
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The solutions lie between -4 and -3; and even though this says "greater than or equal to,"
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I don't want to include an "equal to" that will include an excluded value.
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So, I am going to say that x is greater than -4; I am not going to say greater than or equal to, or I would be including a value that I am not allowed to include.
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So, x is greater than -4 and less than or equal to -3.
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In addition, x is greater than 0; that is also part of the solution set, which is
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any of these values (values for x that are greater than 0, or values of x that are greater than -4 and less than or equal to -3)...
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see, the -3 can be included in the solution set, because it is not an excluded value.
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It is dividing this up, because it was a solution to this equation, not because it was an excluded value.
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As you can see, this is pretty complicated; it takes a lot of steps.
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Your first step is to find the excluded values; I found those.
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The second step is to solve the related equation; I did that right here, and I got that x equals -3.
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This gave me three values that I used to divide up the number line into four intervals.
00:27:49.900 --> 00:27:58.200
My next step was to take test points: I took a test point for A right here, and I found that that did not satisfy the inequality.
00:27:58.200 --> 00:28:05.400
So, this interval is not part of the solution set; A was right here; B was right here.
00:28:05.400 --> 00:28:13.800
x equals -3/2; I solved that out, and this did satisfy the inequality, so this is part of the solution set.
00:28:13.800 --> 00:28:22.500
For C, I took x = -2 and went through; and this did not satisfy the inequality, so this is not part of the solution set.
00:28:22.500 --> 00:28:32.200
And finally, in D, I let x equal 1; that satisfied the inequality, and therefore, the interval D contains part of the solution set.
00:28:32.200 --> 00:28:46.300
So, the solutions are x > -4 and x ≤ -3, and also x > 0 is part of the solution set.
00:28:46.300 --> 00:28:51.500
Let's get some more practice by doing some examples.
00:28:51.500 --> 00:29:01.200
Going back to rational equations, recall that we need to find the LCD.
00:29:01.200 --> 00:29:13.200
And in order to find the LCD, I am going to factor out these denominators; so this one is factored.
00:29:13.200 --> 00:29:21.200
1 - x is already factored; but if we factor out a -1, it is going to make our work easier.
00:29:21.200 --> 00:29:32.100
Recall that we talked earlier about the fact that, when you had two factors that were close, but not exact, factoring out a -1 could help.
00:29:32.100 --> 00:29:40.700
Let's look at this second denominator: I am going to rewrite this, instead of as 1 - x, as -x + 1.
00:29:40.700 --> 00:29:46.300
Now, I can look and see that these two are the same, except the signs of both terms are opposite.
00:29:46.300 --> 00:29:57.100
What this tells me is that, if I factor -1 out of either one (not both--I can pick one or the other and factor out a -1), -1 pulled out of here will give me x.
00:29:57.100 --> 00:30:07.000
-1 pulled out of 1 will give me -1, because, if I multiply this by -1, that is going to give me -x; -1 times -1 will give me a 1 back.
00:30:07.000 --> 00:30:12.500
Now, I can see that I actually have two factors that are the same, plus this -1.
00:30:12.500 --> 00:30:19.800
I could include this 4 as part of the LCD, and then this fraction would end up getting eliminated as well.
00:30:19.800 --> 00:30:24.600
But I am not really worried about this fraction, because it is a constant; I can work with the number 1/4.
00:30:24.600 --> 00:30:32.700
So, you can either include the 4 or not include it when you are using the method of eliminating fractions to solve a rational equation.
00:30:32.700 --> 00:30:37.600
I am actually going to not use that 4, and I am just going to deal with the constant later on.
00:30:37.600 --> 00:30:43.900
But we still need to multiply this term by the LCD that we find here, as well.
00:30:43.900 --> 00:30:54.500
OK, so the LCD is going to be the product of (x - 1), because that is a factor, and -1.
00:30:54.500 --> 00:31:09.800
So, I am going to multiply each term in this equation by this LCD.
00:31:09.800 --> 00:31:20.200
I am also going to rewrite this in its factored form, just to make it simpler to see that what I am working with, this, is the same as this.
00:31:20.200 --> 00:31:33.700
I just factored this into this by factoring out the -1.
00:31:33.700 --> 00:31:37.400
OK, let's go ahead and start canceling common factors.
00:31:37.400 --> 00:31:54.500
The (x - 1)s cancel; this gives me -1 times 3, which is simply -3, minus...here I can cancel out the -1 and the (x - 1)s.
00:31:54.500 --> 00:32:08.800
So, this gives me -4: so -3 - 4 equals -1, times (x - 1), all divided by 4.
00:32:08.800 --> 00:32:14.100
I am not worried about this, because the denominator does not have a variable in it, so it is not a rational expression.
00:32:14.100 --> 00:32:18.100
I have gotten rid of all of my rational expressions: remember, rational expressions are algebraic fractions,
00:32:18.100 --> 00:32:22.800
but we are talking about fractions where there is a variable in the denominator.
00:32:22.800 --> 00:32:39.600
OK, -3 and -4 is -7; equals...let's write this as -x - 1, over 4; therefore, I can multiply both sides of the equation by a -4.
00:32:39.600 --> 00:32:48.100
That will cancel this -4 out, and that gives me -4 times -7, which is 28, equals x - 1.
00:32:48.100 --> 00:32:54.200
I am going to add a 1 to both sides to get 29 = x.
00:32:54.200 --> 00:33:03.600
Now, I can't forget about my excluded values: these are going to be values that make the denominator 0.
00:33:03.600 --> 00:33:11.400
And for this first one, I have (x - 1); if that equals 0, then this will be undefined.
00:33:11.400 --> 00:33:17.600
So, that would occur in cases where x equals 1; so x = 1 is an excluded value.
00:33:17.600 --> 00:33:28.900
Looking over here, I factored this into -1 times (x - 1); using the zero product property, I could again say, if x - 1 equals 0, then I have a problem.
00:33:28.900 --> 00:33:33.300
x = 1 is the excluded value for this one and for this one.
00:33:33.300 --> 00:33:41.400
So, since this solution I got, x = 29, is not an excluded value, then it is a valid solution.
00:33:41.400 --> 00:33:52.100
If I came up with the solution x = 1, that would have been an extraneous solution that I would have had to throw out.
00:33:52.100 --> 00:33:59.900
This example is another rational equation that we are again going to solve by finding the LCD.
00:33:59.900 --> 00:34:04.400
So, let's go ahead and just factor the denominators right here.
00:34:04.400 --> 00:34:13.900
This is x plus a factor, times x minus a factor; factors of 10 are 1 and 10, 2 and 5.
00:34:13.900 --> 00:34:22.000
And I need those to add up to 3x; so 5 - 2 would give me a 3.
00:34:22.000 --> 00:34:38.500
Therefore, the correct factorization would be (x + 5) (x - 2), plus...I am going to leave some room to multiply these by the LCD.
00:34:38.500 --> 00:34:50.700
So, 2/(x - 2) = 19/(x + 5): I have this factored out right here.
00:34:50.700 --> 00:35:07.300
And the LCD is going to be (x + 5); that is present once; and (x - 2)--that is also present once; that is the LCD.
00:35:07.300 --> 00:35:18.500
So, I am going to multiply each term by that LCD, (x + 5) (x - 2),
00:35:18.500 --> 00:35:34.900
plus (x + 5) (x - 2) times this next term, times (x + 5) (x - 2) times this third term.
00:35:34.900 --> 00:35:43.100
Go ahead and cancel out common factors: here, both factors are common, leaving a 3 behind.
00:35:43.100 --> 00:35:57.900
Here, the (x - 2) cancels out, leaving behind 2(x + 5), and on the right, the (x + 5) cancels out; that leaves me with 19(x - 2).
00:35:57.900 --> 00:36:16.400
Let's go ahead and solve this equation: 3 +...2 times x is 2x, plus 2 times 5 is 10, equals 19x...19 times -2 is -38.
00:36:16.400 --> 00:36:24.800
Combining the constants on the left, 2x + 13 = 19x - 38.
00:36:24.800 --> 00:36:37.400
I am going to subtract the 2x, first, from each side: that is going to give me 13 = 17x - 38.
00:36:37.400 --> 00:36:45.300
And then, I am going to go ahead and add a 38 to both sides to give me 51 = 17x.
00:36:45.300 --> 00:36:55.300
Divide both sides by 17; this is going to give me 51/17 = x; therefore, x = 3.
00:36:55.300 --> 00:36:59.400
Now, before I say that this is the actual solution, I need to look for excluded values.
00:36:59.400 --> 00:37:02.900
And it is easy to do that, because I have already factored these out.
00:37:02.900 --> 00:37:10.100
Excluded values: this is something you can do right at the beginning, before you start working, or right when you finish.
00:37:10.100 --> 00:37:13.900
But you have to make sure that you check the solutions, each time, for these.
00:37:13.900 --> 00:37:25.600
So, for this first one, the denominator is (x + 5) (x - 2) = 0...that would then result in a denominator that is 0, which is not allowed.
00:37:25.600 --> 00:37:34.100
So, using the zero product property, if (x + 5) equals 0, this whole thing will equal 0.
00:37:34.100 --> 00:37:41.300
Or if (x - 2) equals 0, this whole denominator will equal 0, when you multiply 0 times the other factor.
00:37:41.300 --> 00:37:48.800
So, x = -5 and x = 2 are excluded values.
00:37:48.800 --> 00:37:56.800
Looking right here, this is the same factor as here; so this also has an excluded value of x = 2; I have that accounted for.
00:37:56.800 --> 00:38:04.400
And here, if x + 5 = 0, that would be the same as this, x = -5; so I don't have to worry about these--they are already accounted for.
00:38:04.400 --> 00:38:09.500
Excluded values for all 3 (I have covered all 3) are these two.
00:38:09.500 --> 00:38:19.300
I look over here, and my solution, x = 3, is not an excluded value; therefore, it is a valid solution.
00:38:19.300 --> 00:38:22.000
OK, again, we have a rational equation we need to solve.
00:38:22.000 --> 00:38:28.600
And the first step is to find the LCD by factoring out the denominators.
00:38:28.600 --> 00:38:37.800
So, this first denominator (we will find the LCD)...I just have x + 1; that is already factored.
00:38:37.800 --> 00:38:45.300
x² - 1 is the difference of two squares, so that is (x + 1) (x - 1).
00:38:45.300 --> 00:38:55.300
Therefore, the LCD...I have this (x + 1) factor here and here, and I have an x - 1.
00:38:55.300 --> 00:39:11.200
So, I am going to solve this by multiplying each term by (x + 1) (x - 1)...times x, equals (x + 1) (x - 1)
00:39:11.200 --> 00:39:28.200
times (x² + x + 2)/(x + 1), minus (x + 1) (x - 1) times (x² - 5), divided by...
00:39:28.200 --> 00:39:36.100
I am going to rewrite this in the factored form, so it is more obvious what cancels and what doesn't.
00:39:36.100 --> 00:39:45.900
All right, so over here, I end up with just x, times (x + 1) (x - 1).
00:39:45.900 --> 00:39:57.400
Here, the (x + 1)s cancel out; that is going to leave me with (x - 1) times (x² + x + 2),
00:39:57.400 --> 00:40:05.800
minus...the (x + 1)s cancel; the (x - 1)s cancel; so minus (x² - 5).
00:40:05.800 --> 00:40:16.300
Multiplying this out, recall that (x + 1) (x - 1) is (x² - 1), so this gives me x times (x² - 1).
00:40:16.300 --> 00:40:25.200
equals...x times x² is x³; x times x is x²; x times 2 is 2x.
00:40:25.200 --> 00:40:34.800
Multiplying -1 times each of these terms: -x² - x...-1 times 2 is -2.
00:40:34.800 --> 00:40:41.500
A negative times a positive gives me -x²; a negative times a negative gives me + 5.
00:40:41.500 --> 00:40:56.800
Now, to take care of this: this is x³ - x.
00:40:56.800 --> 00:41:04.400
All right, first let's take care of the x³; I am going to subtract x³ from both sides to move this from the right to the left.
00:41:04.400 --> 00:41:09.700
So, what happens is: I do x³ - x³; this drops out.
00:41:09.700 --> 00:41:15.100
I have to do the same thing to the left: x³ - x³--that drops out.
00:41:15.100 --> 00:41:20.100
So, that took care of the x³; this looked worse than it actually was.
00:41:20.100 --> 00:41:24.700
It is leaving me with -x =...well, let's see if I can do any simplifying here on the right.
00:41:24.700 --> 00:41:41.400
I have x² and -x², so those cancel out.
00:41:41.400 --> 00:42:04.500
That leaves me with 2x...so this is gone; this is gone; this is gone...minus x, minus 2, minus x², plus 5.
00:42:04.500 --> 00:42:12.000
I still have some work to do; but I am going to see that I can combine some of this still,
00:42:12.000 --> 00:42:32.700
which gives me -x = 2x - x (that is x), and combining the constants is going to give me -2 + 5 (that is positive 3), minus x².
00:42:32.700 --> 00:42:37.800
Now, what I am going to end up with here is a quadratic equation that I need to solve.
00:42:37.800 --> 00:42:41.000
Let's finish doing the simplification, and then go ahead and solve that.
00:42:41.000 --> 00:42:52.200
I am going to move this x over to the right by adding an x to both sides; that is going to leave me with 0 = 2x + 3 - x².
00:42:52.200 --> 00:42:57.100
I want to write this in a more standard form, where the x² is going to be positive.
00:42:57.100 --> 00:43:05.800
So, let's go ahead and flip the sides, which will give me x² - 2x - 3 = 0.
00:43:05.800 --> 00:43:18.100
All I did is added an x² to both sides, subtracted a 3 from both sides, and subtracted a 2x from both sides, in order to flip this around.
00:43:18.100 --> 00:43:22.600
OK, now that I have just a quadratic equation left, all I need to do is solve it.
00:43:22.600 --> 00:43:34.600
And I can use factoring: this is negative, so I have (x + a factor), times (x - a factor), equals 0.
00:43:34.600 --> 00:43:44.300
Factors of 3 are 1 and 3; and I want them to add up to a negative number, so I am going to make the larger factor negative and the smaller factor positive.
00:43:44.300 --> 00:43:54.700
And this does work; you get x²...the outer term is -3x...plus x; that gives a -2x for the middle term; 1 times -3 is -3.
00:43:54.700 --> 00:44:00.500
That is factored correctly; now I just need to use the zero product property to solve.
00:44:00.500 --> 00:44:10.200
x + 1 = 0; therefore, x = -1; the other solution is that x - 3 = 0; so x = 3.
00:44:10.200 --> 00:44:18.600
The two solutions that I have are possible solutions; and I say "possible," because I have to make sure that these are not excluded values.
00:44:18.600 --> 00:44:24.800
Possible solutions are x = -1 and x = 3.
00:44:24.800 --> 00:44:39.600
Now, let's look for excluded values: excluded values are going to be any values that make either of the denominators 0.
00:44:39.600 --> 00:44:45.200
Here, I have x² - 1; that factored into (x + 1) (x - 1).
00:44:45.200 --> 00:44:50.600
So, if (x + 1) (x - 1) equals 0, the denominator becomes 0.
00:44:50.600 --> 00:44:56.100
So, any values of x that result in this being 0 are excluded.
00:44:56.100 --> 00:45:11.200
Using the zero product property: x + 1 = 0; x - 1 = 0--if either of those occurs, this entire thing will be 0, and you would have an invalid situation.
00:45:11.200 --> 00:45:19.200
Therefore, for this first one, if x equals -1, this whole thing will become 0,
00:45:19.200 --> 00:45:23.700
because -1 times -1 is 1, minus 1--this would give you a 0.
00:45:23.700 --> 00:45:29.200
The other excluded value is x = 1, because 1² is 1, minus 1 would give you 0.
00:45:29.200 --> 00:45:37.400
So, I have these two excluded values: looking over here, I already took care of that, because this is a factor, right here.
00:45:37.400 --> 00:45:43.600
So, I already said that, if x + 1 equals 0, x equals -1, and that is excluded; so that one is already covered.
00:45:43.600 --> 00:45:49.300
Now, I need to compare my solutions to my possible solutions, to these.
00:45:49.300 --> 00:45:58.300
And I see that I only have one valid solution, because this is actually not a valid solution.
00:45:58.300 --> 00:46:03.800
It is an excluded value; I can't include that as part of the solution--it is an extraneous solution.
00:46:03.800 --> 00:46:07.900
x = 3 is the only valid solution.
00:46:07.900 --> 00:46:17.100
This took quite a few steps: the first step was to find the LCD and multiply each term by the LCD to get rid of the fractions.
00:46:17.100 --> 00:46:22.700
And then, the hardest work was actually just multiplying all of this out, keeping track of all the signs,
00:46:22.700 --> 00:46:28.600
and then getting down to where we had a quadratic equation that we solve by factoring to get two possible solutions.
00:46:28.600 --> 00:46:36.100
The final step was to find the excluded values, based on setting the denominators equal to 0 and then solving for x,
00:46:36.100 --> 00:46:44.200
and then comparing my possible solutions against these excluded values, thus eliminating x = -1 from the solution set,
00:46:44.200 --> 00:46:49.400
and being left with one solution, which is x = 3.
00:46:49.400 --> 00:46:54.400
In Example 4, we are going to be working with a rational inequality.
00:46:54.400 --> 00:46:58.700
Recall that, for rational inequalities, we are going to find the excluded values.
00:46:58.700 --> 00:47:04.400
We are going to solve the related equation; and then we are going to use those values to divide the number line into intervals,
00:47:04.400 --> 00:47:12.400
and then test each interval, using a test value to determine where the solution set is.
00:47:12.400 --> 00:47:25.800
Let's first find excluded values: these are going to be values for which the denominator becomes 0.
00:47:25.800 --> 00:47:32.800
If 2x equals 0, that would occur when x equals 0/2, or x equals 0.
00:47:32.800 --> 00:47:38.600
The same thing is going to occur here with 4x--that if x equals 0, it will be an excluded value.
00:47:38.600 --> 00:47:42.100
So, the excluded value is x = 0.
00:47:42.100 --> 00:47:53.900
Now, I need to solve the related equation, 1/2x + 5/4x - 1 = 0.
00:47:53.900 --> 00:48:07.600
And the least common denominator of 2x and 4x...I could factor this out to 2 times 2.
00:48:07.600 --> 00:48:17.700
So, the LCD is going to be...there is a 2 present once here, but twice here; and remember, I take each unique factor
00:48:17.700 --> 00:48:22.400
to the highest power that it is present, so I am going to say 2².
00:48:22.400 --> 00:48:27.400
There is an x present once here and once here, so it is 2² times x equals 4x.
00:48:27.400 --> 00:48:47.800
So, the LCD is 4x; so multiplying here, 4x(1/2x) + 4x(5/4x) - 1(4x) = 0(4x).
00:48:47.800 --> 00:48:55.000
Canceling out common factors, this becomes a 2; 2 cancels out; the x's cancel out.
00:48:55.000 --> 00:49:09.200
2 times 1 is 2, plus...the 4x's cancel, leaving behind 5; this is -4x = 0.
00:49:09.200 --> 00:49:28.200
7 - 4x = 0; 7 = 4x; divide both sides by 4; that gives x = 7/4, or this could be rewritten as x = 1 3/4.
00:49:28.200 --> 00:49:34.400
So, that is my excluded value; and my solution to the related equation is x = 1 3/4.
00:49:34.400 --> 00:49:44.100
Let's go down here, and I am going to use this value and this value to divide up my number line.
00:49:44.100 --> 00:49:52.400
I have 0 here, and I have 1 3/4 here.
00:49:52.400 --> 00:50:03.500
So, I end up with three intervals: A, B, and C; and I need to test a point within each of these intervals.
00:50:03.500 --> 00:50:10.500
And if that value satisfies the inequality, I know that this interval is at least part of the solution set.
00:50:10.500 --> 00:50:22.000
For the first interval (for A), I am going to let x equal -1; I need something less than 0; I will let x equal -1--that is simple to work with.
00:50:22.000 --> 00:50:31.200
That is going to give me 1 over 2(-1), plus 5 divided by 4(-1), minus 1, is greater than 0.
00:50:31.200 --> 00:50:40.200
And let's see if this holds up: this gives me -1/2, minus 5/4, minus 1, is greater than 0.
00:50:40.200 --> 00:50:49.400
This is the same as saying -2/4 (just converting to a common denominator) - 5/4 - 1 > 0.
00:50:49.400 --> 00:50:52.700
I actually don't even need to go farther; in fact, I even kind of just looked at this and said,
00:50:52.700 --> 00:50:57.100
"I have a bunch of negatives; those are not going to be greater than 0 when I combine them."
00:50:57.100 --> 00:51:08.800
So, no: this is not valid; therefore, that interval does not contain the solution set or part of the solution set.
00:51:08.800 --> 00:51:16.900
For B, I have values between 0 and 1 3/4, so I can use 1 as a test point, x = 1.
00:51:16.900 --> 00:51:26.200
This is going to give me 1 over 2(1), plus 5 divided by 4(1), minus 1, is greater than 0,
00:51:26.200 --> 00:51:41.300
which is 1/2 + 5/4 - 1 > 0; I can just convert this to 2/4 + 5/4 - 1 > 0.
00:51:41.300 --> 00:51:55.900
And this is going to give me 5/4 + 2/4, is going to give me 7/4: 7/4 is greater than 1, so when I subtract 1 from 7/4, I am going to end up with 3/4.
00:51:55.900 --> 00:51:59.300
So, this is going to give me 3/4 > 0.
00:51:59.300 --> 00:52:04.700
Even if I didn't figure this whole thing out, as soon as I saw that this is a positive number larger than 1,
00:52:04.700 --> 00:52:09.600
I know that, when I subtract 1 from it, I will get something positive; so it is going to be greater than 0.
00:52:09.600 --> 00:52:17.700
So, this one is valid; therefore, yes for B: this interval does contain at least part of the solution set.
00:52:17.700 --> 00:52:28.600
For C, I am going to go right here, and I am going to use 2 as my test point: x = 2.
00:52:28.600 --> 00:52:36.200
This is going to give me 1 over 2(2), plus 5 over 4(2), minus 1, is greater than 0.
00:52:36.200 --> 00:52:43.200
That is 1/4 + 5/8 - 1 > 0.
00:52:43.200 --> 00:52:52.200
I need to find a common denominator here, so I am going to convert 1/4 to 2/8: plus 5/8, minus 1, is greater than 0.
00:52:52.200 --> 00:53:03.000
That is 7/8 - 1 > 0; well, 7/8 - 1 is going to leave -1/8 > 0.
00:53:03.000 --> 00:53:08.400
And that is not true, so C is not part of the solution set.
00:53:08.400 --> 00:53:16.400
Therefore, the solution for this...the possible solution...I am going to say "possible,"
00:53:16.400 --> 00:53:29.400
because we have to look back at excluded values...is x is greater than 0, but it is less than 1 3/4.
00:53:29.400 --> 00:53:43.400
Now, let's look back up at excluded values: I cannot let x equal 0--that is an excluded value.
00:53:43.400 --> 00:53:53.700
But I am OK here, because x is greater than 0; so I am covered, and this is my actual solution--this is valid.
00:53:53.700 --> 00:54:07.200
And just looking back to review: find your excluded values (x = 0 is an excluded value), then solve the related rational equation.
00:54:07.200 --> 00:54:13.500
I did that by multiplying by the LCD, multiplying each term, getting rid of those fractions,
00:54:13.500 --> 00:54:22.200
then finding that x = 1 3/4 was the solution, and it wasn't an excluded value (so it was a valid solution).
00:54:22.200 --> 00:54:28.000
I used those two values, 0 and 1 3/4, to divide the number line into intervals.
00:54:28.000 --> 00:54:36.600
And then, I tested each interval: I tested interval A and found that my test value did not satisfy the inequality, so that is not part of the solution set.
00:54:36.600 --> 00:54:43.800
I tested B, using x = 1, and found the inequality did hold up, so B contains part of my solution set.
00:54:43.800 --> 00:54:49.300
C--I tested: that value did not satisfy the inequality, so that is not part of the solution set.
00:54:49.300 --> 00:54:53.600
Therefore, the solution set is that x is greater than 0 and less than 1 3/4.
00:54:53.600 --> 00:55:00.200
But remember: if you are working with greater than or equal to in the original, and you are thinking,
00:55:00.200 --> 00:55:07.400
"OK, I am going to put an 'equal to' here," you need to be careful that you don't encompass an excluded value as part of that.
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That concludes this lesson of Educator.com on rational equations and inequalities.
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I will see you next lesson!