## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Solving Rational Equations and Inequalities

- To solve a rational equation, multiply each term on both sides by the LCD of all the denominators in the equation. Then solve the resulting equation, which has no fractions.
- Always check for extraneous solutions – values that make one or more of the denominators equal to 0. Exclude such values from the solution set. Better yet, before solving the equation, determine the values that must be excluded by setting each denominator equal to 0 and solving. Then you will recognize an extraneous solution as soon as it appears.

### Solving Rational Equations and Inequalities

- List the prime factors of 3n + 6 and 3 to find the LCD. Factoring may be required.
- 3 =
- 3n + 6 =
- 3 = 3
- 3n + 6 = 3(n + 2)
- Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = 3(n + 2)
- Change each rational expression into an equivalent expression with the LCD.
- [5/3] − [1/3] = [(n + 1)/(3n + 6)] → [4/3] = [(n + 1)/(3n + 6)]
- [4/3]( [(3(n + 2))/(3(n + 2))] ) = [(n + 1)/(3(n + 2))]
- [4/]( [((n + 2))/(3(n + 2))] ) = [(n + 1)/(3(n + 2))]
- [(4(n + 2))/(3(n + 2))] = [((n + 1))/(3(n + 2))]
- Cancel out (n + 2)
- [4/3] = [((n + 1))/(3(n + 2))]
- Cross Multiply
- 3(n + 1) = 12(n + 2)
- 3n + 3 = 12n + 24
- − 21 = 9n

^{2}− 14x + 12)] + [5/(2x − 2)] = [1/(x − 6)]

- List the prime factors of 2x
^{2}− 14x + 12, 2x − 2 and x − 6 to find the LCD. Factoring may be required. - 2x
^{2}− 14x + 12 = - 2x − 2 =
- x − 6 =
- 2x
^{2}− 14x + 12 = 2*(x − 6)(x − 1) - 2x − 2 = 2*(x − 1)
- x − 6 = (x − 6)
- Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = 2*(x − 6)(x − 1)
- Change each rational expression into an equivalent expression with the LCD.
- [1/(2x
^{2}− 14x + 12)] + [5/(2x − 2)] = [1/(x − 6)] → [1/(2(x − 1)(x − 6))] + [5/(2(x − 1))] = [1/(x − 6)] - [1/(2(x − 1)(x − 6))] + [5/(2(x − 1))]( [((x − 6))/((x − 6))] ) = [1/(x − 6)]( [(2(x − 1))/(2(x − 1))] )
- [1/(2(x − 1)(x − 6))] + [(5(x − 6))/(2(x − 1)(x − 6))] = [(2(x − 1))/(2(x − 1)x − 6)]
- Add the left side of the equation
- [(1 + 5(x − 6))/(2(x − 1)(x − 6))] = [(2(x − 1))/(2(x − 1)(x − 6))]
- Subtract the right side of the eqution
- [(5x − 29)/(2(x − 1)(x − 6))] − [(2(x − 1))/(2(x − 1)(x − 6))] = 0
- [(5x − 29)/(2(x − 1)(x − 6))] − [(2x − 2)/(2(x − 1)(x − 6))] = 0
- Subtract
- [((5x − 29) − (2x − 2))/(2(x − 1)(x − 6))] = 0 → [((5x − 29) − 2x + 2))/(2(x − 1)(x − 6))] = 0 → [(3x − 27)/(2(x − 1)(x − 6))] = 0
- [(3x − 27)/(2(x − 1)(x − 6))] = 0
- Multiply both sides of the equation by the denominator, simplify
- ( 2(x − 1)(x − 6) )[(3x − 27)/(2(x − 1)(x − 6))] = 0(2(x − 1)(x − 6))
- 3x − 27 = 0
- 3x = 27

- List the prime factors of x − 6;3x − 18 and 3 to find the LCD. Factoring may be required.
- x − 6 =
- 3x − 18 =
- 3 =
- x − 6 = (x − 6)
- 3x − 18 = 3(x − 6)
- 3 = 3
- Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = 3(x − 6)
- Change each rational expression into an equivalent expression with the LCD.
- [x/(x − 6)] − [1/(3x − 18)] = [1/3] → [x/(x − 6)] − [1/(3(x − 6))] = [1/3]
- [x/(x − 6)]( [3/3] ) − [1/(3(x − 6))] = [1/3]( [(x − 6)/(x − 6)] )
- [3x/(3(x − 6))] − [1/(3(x − 6))] = [((x − 6))/(3(x − 6))]
- Add the left side of the equation
- [(3x − 1)/(3(x − 6))] = [((x − 6))/(3(x − 6))]
- Subtract the right side of the eqution
- [(3x − 1)/(3(x − 6))] − [((x − 6))/(3(x − 6))] = 0
- Subtract
- [((3x − 1) − (x − 6))/(3(x − 6))] = 0 → [((3x − 1) − x + 6)/(3(x − 6))] = 0 → [(2x + 5)/(3(x − 6))] = 0
- [(2x + 5)/(3(x − 6))] = 0
- Multiply both sides of the equation by the denominator, simplify
- ( 3(x − 6) )[(2x + 5)/(3(x − 6))] = 0(3(x − 6))
- 2x + 5 = 0
- 2x = − 5

^{2}− 9x + 10)]

- List the prime factors of 2x − 5;x − 2 and 2x
^{2}− 9x + 10 to find the LCD. Factoring may be required. - 2x − 5 =
- x − 2 =
- 2x
^{2}− 9x + 10 = - 2x − 5 = (2x − 5)
- x − 2 = (x − 2)
- 2x
^{2}− 9x + 10 = (2x − 5)(x − 2) - Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = (2x − 5)(x − 2)
- Change each rational expression into an equivalent expression with the LCD.
- [1/(2x − 5)] − [1/(x − 2)] = [(x + 5)/(2x
^{2}− 9x + 10)] → [1/(2x − 5)] − [1/(x − 2)] = [(x + 5)/((2x − 5)(x − 2))] - [1/(2x − 5)]( [(x − 2)/(x − 2)] ) − [1/(x − 2)]( [(2x − 5)/(2x − 5)] ) = [(x + 5)/((2x − 5)(x − 2))]
- [((x − 2))/((2x − 5)(x − 2))] − [((2x − 5))/((x − 2)(2x − 5))] = [(x + 5)/((2x − 5)(x − 2))]
- Subtract the left side of the equation
- [((x − 2) − (2x − 5))/((2x − 5)(x − 2))] = [(x + 5)/((2x − 5)(x − 2))]
- [(x − 2 − 2x + 5)/((2x − 5)(x − 2))] = [(x + 5)/((2x − 5)(x − 2))]
- [(3 − x)/((2x − 5)(x − 2))] = [(x + 5)/((2x − 5)(x − 2))]
- Subtract the right side of the eqution
- [(3 − x)/((2x − 5)(x − 2))] − [(x + 5)/((2x − 5)(x − 2))] = 0
- Subtract
- [(3 − x − (x + 5))/((2x − 5)(x − 2))] = 0 → [(3 − x − x − 5)/((2x − 5)(x − 2))] = 0 → [( − 2x − 2)/((2x − 5)(x − 2))] = 0
- [( − 2x − 2)/((2x − 5)(x − 2))] = 0
- Multiply both sides of the equation by the denominator, simplify
- ( (2x − 5)(x − 2) )[( − 2x − 2)/((2x − 5)(x − 2))] = 0((2x − 5)(x − 2))
- − 2x − 2 = 0
- − 2x = 2

^{2}+ x)] = [1/(x + 1)]

- List the prime factors of x, x
^{2}+ x and x + 1 to find the LCD. Factoring may be required. - x =
- x
^{2}+ x = - x + 1 =
- x = x
- x
^{2}+ x = x(x + 1) - x + 1 = (x + 1)
- Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = x(x + 1)
- Change each rational expression into an equivalent expression with the LCD.
- [3/x] + [1/(x
^{2}+ x)] = [1/(x + 1)] → [3/x] + [1/(x(x + 1))] = [1/(x + 1)] - [3/x]( [((x + 1))/((x + 1))] ) + [1/(x(x + 1))] = [1/((x + 1))]( [x/x] )
- [(3(x + 1))/(x(x + 1))] + [1/(x(x + 1))] = [x/(x(x + 1))]
- Add the left side of the equation
- [(3(x + 1) + 1)/(x(x + 1))] = [x/(x(x + 1))]
- [(3x + 4)/(x(x + 1))] = [x/(x(x + 1))]
- Subtract the right side of the eqution
- [(3x + 4)/(x(x + 1))] − [x/(x(x + 1))] = 0
- Subtract
- [(3x + 4 − x)/(x(x + 1))] = 0
- [(2x + 4)/(x(x + 1))] = 0
- Multiply both sides of the equation by the denominator, simplify
- ( x(x + 1) )[(2x + 4)/(x(x + 1))] = 0((x(x + 1))
- 2x + 4 = 0
- 2x = − 4

^{2}− 6x)] = [1/(x − 6)]

- List the prime factors of x, x
^{2}− 6x and x − 6 to find the LCD. Factoring may be required. - x =
- x
^{2}− 6x = - x − 6 =
- x = x
- x
^{2}− 6x = x(x − 6) - x − 6 = (x − 6)
- Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = x(x − 6)
- Change each rational expression into an equivalent expression with the LCD.
- [6/x] − [1/(x
^{2}− 6x)] = [1/(x − 6)] → [6/x] − [1/(x(x − 6))] = [1/(x − 6)] - [6/x]( [((x − 6))/((x − 6))] ) − [1/(x(x − 6))] = [1/(x − 6)]( [x/x] )
- [(6(x − 6))/(x(x − 6))] − [1/(x(x − 6))] = [x/(x(x − 6))]
- Add the left side of the equation
- [(6(x − 6) − 1)/(x(x − 6))] = [x/(x(x − 6))]
- [(6x − 37)/(x(x − 6))] = [x/(x(x − 6))]
- Subtract the right side of the eqution
- [(6x − 37)/(x(x − 6))] − [x/(x(x − 6))] = 0
- Subtract
- [(6x − 37 − x)/(x(x − 6))] = 0
- [(5x − 37)/(x(x − 6))] = 0
- Multiply both sides of the equation by the denominator, simplify
- ( x(x − 6) )[(5x − 37)/(x(x − 6))] = 0((x(x − 6))
- 5x − 37 = 0
- 5x = 37

^{2}− 1)] = 1

- List the prime factors of x − 1, x
^{2}− 1 to find the LCD. Factoring may be required. - x − 1 =
- x
^{2}− 1 = - x − 1 = (x − 1)
- x
^{2}− 1 = (x − 1)(x + 1) - Use each prime factor the greatest number of times it appears in each of the factorization
- LCD =
- LCD = (x + 1)(x − 1)
- Change each rational expression into an equivalent expression with the LCD.
- [(x − 6)/(x − 1)] + [4/(x
^{2}− 1)] = 1 - [(x − 6)/(x − 1)]( [(x + 1)/(x + 1)] ) + [4/((x − 1)(x + 1))] = 1( [((x − 1)(x + 1))/((x − 1)(x + 1))] )
- [((x − 6)(x + 1))/((x − 1)(x + 1))] + [4/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
- Add the left side of the equation
- [((x − 6)(x + 1))/((x − 1)(x + 1))] + [4/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
- [((x − 6)(x + 1) + 4)/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
- [(x
^{2}− 5x − 2)/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))] - Subtract the right side of the eqution
- [(x
^{2}− 5x − 2)/((x − 1)(x + 1))] − [((x − 1)(x + 1))/((x − 1)(x + 1))] = 0 - Subtract
- [(x
^{2}− 5x − 2 − (x^{2}− 1))/((x − 1)(x + 1))] = 0 - [( − 5x − 1)/((x − 1)(x + 1))] = 0
- Multiply both sides of the equation by the denominator, simplify
- ( (x − 1)(x + 1) )[( − 5x − 1)/((x − 1)(x + 1))] = 0((x − 1)(x + 1))
- − 5x − 1 = 0
- − 5x = 1

- Identify excluded values
- x − 6 = 0
- x = 6
- 3x − 18 = 0
- 3x = 18
- x = 6
- You cannot have 6 as part of the solution set
- Solve the related equation
- [x/(x − 6)] − [1/(3x − 18)] = [1/3]
- [x/(x − 6)] − [1/(3x − 18)] = [1/3] → [x/(x − 6)] − [1/(3(x − 6))] = [1/3]
- [x/(x − 6)]( [3/3] ) − [1/(3(x − 6))] = [1/3]( [(x − 6)/(x − 6)] )
- [3x/(3(x − 6))] − [1/(3(x − 6))] = [((x − 6))/(3(x − 6))]
- Add the left side of the equation
- [(3x − 1)/(3(x − 6))] = [((x − 6))/(3(x − 6))]
- Subtract the right side of the eqution
- [(3x − 1)/(3(x − 6))] − [((x − 6))/(3(x − 6))] = 0
- Subtract
- [((3x − 1) − (x − 6))/(3(x − 6))] = 0 → [((3x − 1) − x + 6)/(3(x − 6))] = 0 → [(2x + 5)/(3(x − 6))] = 0
- [(2x + 5)/(3(x − 6))] = 0
- Multiply both sides of the equation by the denominator, simplify
- ( 3(x − 6) )[(2x + 5)/(3(x − 6))] = 0(3(x − 6))
- 2x + 5 = 0
- 2x = − 5
- x = − [5/2]
- Test different intervals to check for the solution set
- Test: x=−4

[x/(x − 6)] − [1/(3x − 18)] < [1/3]

[( − 4)/( − 4 − 6)] − [1/(3( − 4) − 18)] < [1/3]

[4/10] + [1/20] < [1/3]

Not True - Test: x=0

[x/(x − 6)] − [1/(3x − 18)] < [1/3]

[0/(0 − 6)] − [1/(3(0) − 18)] < [1/3]

[1/18] < [1/3]

True - Test: x=10

[x/(x − 6)] − [1/(3x − 18)] < [1/3]

[10/(10 − 6)] − [1/(3(10) − 18)] < [1/3]

[10/4] − [1/12] < [1/3]

Not True - Therefore, the solution is

^{2}− 1)] ≤ 1

- Identify excluded values
- x − 1 = 0
- x = 1
- x
^{2}− 1 = 0 - x = 1;x = − 1
- You cannot have 1 and − 1 as part of the solution set
- Solve the related equation
- [(x − 6)/(x − 1)] + [4/(x
^{2}− 1)] = 1 - [(x − 6)/(x − 1)]( [(x + 1)/(x + 1)] ) + [4/((x − 1)(x + 1))] = 1( [((x − 1)(x + 1))/((x − 1)(x + 1))] )
- [((x − 6)(x + 1))/((x − 1)(x + 1))] + [4/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
- Add the left side of the equation
- [((x − 6)(x + 1))/((x − 1)(x + 1))] + [4/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
- [((x − 6)(x + 1) + 4)/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
- [(x
^{2}− 5x − 2)/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))] - Subtract the right side of the eqution
- [(x
^{2}− 5x − 2)/((x − 1)(x + 1))] − [((x − 1)(x + 1))/((x − 1)(x + 1))] = 0 - Subtract
- [(x
^{2}− 5x − 2 − (x^{2}− 1))/((x − 1)(x + 1))] = 0 - [( − 5x − 1)/((x − 1)(x + 1))] = 0
- Multiply both sides of the equation by the denominator, simplify
- ( (x − 1)(x + 1) )[( − 5x − 1)/((x − 1)(x + 1))] = 0((x − 1)(x + 1))
- − 5x − 1 = 0
- − 5x = 1
- x = − [1/5]
- Test different intervals to check for the solution set
- Test: x=−2

[(x − 6)/(x − 1)] + [4/(x^{2}− 1)] ≤ 1

[( − 2 − 6)/( − 2 − 1)] + [4/(( − 2)^{2}− 1)] ≤ 1

[8/3] + [4/3] ≤ 1

Not True - Test: x=−[1/2]

[(x − 6)/(x − 1)] + [4/(x^{2}− 1)] ≤ 1

[( − [1/2] − 6)/( − [1/2] − 1)] + [4/(( − [1/2])^{2}− 1)] ≤ 1

[13/3] − [16/3] ≤ 1

True - Test: x=0

[(x − 6)/(x − 1)] + [4/(x^{2}− 1)] ≤ 1

[(0 − 6)/(0 − 1)] + [4/(0^{2}− 1)] ≤ 1

6 − 4 ≤ 1

Not True - Test: x=2

[(x − 6)/(x − 1)] + [4/(x^{2}− 1)] ≤ 1

[(2 − 6)/(2 − 1)] + [4/(2^{2}− 1)] ≤ 1

− 4 + [4/3] ≤ 1

True - Therefore, the solution is

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Solving Rational Equations and Inequalities

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Rational Equations
- Example: Algebraic Fraction
- Least Common Denominator
- Example: Simple Rational Equation
- Example: Solve Rational Equation
- Extraneous Solutions
- Rational Inequalities
- Excluded Values
- Solve Related Equation
- Find Intervals
- Use Test Values
- Example: Rational Inequality
- Example: Rational Inequality 2
- Example 1: Rational Equation
- Example 2: Rational Equation
- Example 3: Rational Equation
- Example 4: Rational Inequality

- Intro 0:00
- Rational Equations 0:15
- Example: Algebraic Fraction
- Least Common Denominator
- Example: Simple Rational Equation
- Example: Solve Rational Equation
- Extraneous Solutions 9:31
- Doublecheck
- No Solution
- Example: Extraneous
- Rational Inequalities 14:01
- Excluded Values
- Solve Related Equation
- Find Intervals
- Use Test Values
- Example: Rational Inequality
- Example: Rational Inequality 2
- Example 1: Rational Equation 28:50
- Example 2: Rational Equation 33:51
- Example 3: Rational Equation 38:19
- Example 4: Rational Inequality 46:49

### Algebra 2

### Transcription: Solving Rational Equations and Inequalities

*Welcome to Educator.com.*0000

*In today's lesson, we are going to continue on with talking about rational expressions.*0003

*this time extending our discussion to solving rational equations and rational inequalities.*0007

*The definition of a rational equation is an equation that contains one or more rational expressions.*0016

*And remember that rational expressions are algebraic fractions.*0022

*For example, a rational equation could be something like this: (x ^{2} - 3x + 9)/(x^{2} - 36) + 3/(x + 5) = 1/2.*0027

*The technique for solving rational equations is going to be to eliminate the fractions.*0046

*And that can be done by multiplying each term of the equation by the least common denominator of all the fractions in the equation.*0051

*We are going to find the least common denominator, and then multiply each term in the equation by that expression.*0059

*Recall, from previous lectures: we discussed the least common denominator.*0070

*And, if necessary, go back and review that, because we are going to be applying that idea to solve rational equations.*0074

*Let's look at another, simpler example, and then go ahead and solve it.*0082

*2 divided by (x + 3), minus 1/2x, equals 1/x.*0087

*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.*0095

*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.*0106

*Therefore, if I look at the denominators, they are already factored out; that makes this very simple.*0120

*I have (x + 3), 2x, and x; therefore, the LCD would be (x + 3)...that is a unique factor; in this second denominator,*0127

*I have 2 times x; so 2 and x are both factors; and then I have x--however, x is already represented here.*0140

*And I only need to represent each unique factor the highest number of times that it is present in any one of the denominators.*0152

*So, x is present here once, and it is present here once; so I only have to represent it here once.*0163

*If I had x ^{2}, for example, here, then I would have written this as 2x^{2}(x + 3).*0167

*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.*0174

*And then, we will see what happens.*0182

*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).*0184

*Let's start canceling out to simplify: the (x + 3)'s cancel here, leaving me with 2x times 2, minus...*0212

*here, the 2x's cancel out, leaving me with (x + 3) times 1, or just (x + 3), equals...*0222

*here the x's cancel out, leaving 2 times (x + 3).*0232

*At this point, you can see that the fractions have been eliminated, which was exactly my goal.*0239

*So now, I am just going to set about simplifying this.*0244

*2x times 2 is 4x; minus x, minus 3, equals 2x + 6.*0247

*All I have here now is a linear equation, and I just need to solve for x.*0257

*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;*0261

*and at the same time, I am going to add 3 to both sides; so moving the variables to the left,*0274

*and the constants to the right, gives me x = 9.*0279

*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.*0287

*And we will talk more about values that would cause the denominator to be 0 in a few minutes.*0300

*But for right now, I can see that 9 ^{2} - 36 would not give me 0, and 9 + 5 would not give me 0.*0304

*So, this is a valid solution.*0311

*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.*0314

*Now, it actually would have been possible to solve this by converting all of these rational expressions,*0323

*all of these fractions, to a common denominator, and then just adding and subtracting.*0330

*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.*0334

*Let's do another example to illustrate this technique.*0341

*5/(x ^{2} + x - 2) + 4/(x - 1) = 1/(x + 2).*0346

*Recall that the first step is to find the LCD.*0357

*In order to do that, I am going to go up here and factor out each of the denominators.*0360

*x ^{2} + x - 2: this is going to factor to (x + something) (x - something).*0366

*And I want factors of 2 that add up to the coefficient of 1 in front of the x.*0375

*Therefore, I am going to put the 2 here and 1 here; and this will give me x ^{2} - x + 2x, which will give me x;*0380

*and then, the last term is - 2; so that checks out.*0392

*x - 1 just stays as x - 1; and x + 2 is x + 2.*0397

*So, the LCD: I have this factor, x + 2, and the highest number of times it is present in any one denominator,*0403

*in any one polynomial, is once, so I can leave it as (x + 2).*0410

*(x - 1) is also present only once in each of these two, so I am going to represent it once.*0416

*The LCD is (x + 2) (x - 1).*0422

*I am rewriting this with the denominators in factored form, (x + 2) (x - 1), and leaving some room to multiply by the LCD.*0426

*4/(x - 1) = 1/(x + 2).*0440

*I am going to multiply each term by the LCD: (x + 2) (x - 1).*0448

*Again, here, I am multiplying this by (x + 2) (x - 1); and finally, on the right side, (x + 2) (x - 1).*0456

*The goal is to get rid of the fractions, so let's check out if that worked.*0469

*These cancel; both factors cancel, leaving behind only a 5.*0474

*Right here, the (x - 1)s cancel, leaving behind 4 times (x + 2).*0478

*On the right side, the (x + 2)s cancel, leaving behind 1 times (x - 1), or simply (x - 1).*0486

*Now, I am left with a linear equation that I can solve by isolating x.*0494

*5 + (4x + 8) = x - 1.*0498

*Combining the 8 and the 5 is going to give me 4x + 13 = x - 1.*0505

*I am going to subtract x from both sides to combine the variables on the left, so 4x - x is 3x + 13 = -1.*0512

*Now, I am going to go ahead and subtract 13 from both sides to give me 3x = -14.*0526

*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.*0539

*So again, the technique to solve a rational equation is to find the least common denominator,*0553

*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.*0559

*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.*0572

*Now, we are going to talk about that in detail.*0578

*You might recall that we talked about extraneous solutions when we talked about radical equations*0580

*(equations containing radicals that had a variable as part of the radicand).*0585

*And we used a technique there where we squared both sides of the equation.*0590

*But the result could be the introduction of solutions that were not valid--extraneous solutions.*0594

*Here, the same thing can happen: the technique that we are using will find a solution if one exists.*0600

*But extra solutions can also pop up that are not valid.*0607

*So, using this technique, you might find that you have one extraneous solution and one valid solution;*0613

*all valid solutions; all extraneous solutions; it could be any combination--you just have to check.*0620

*And if you find that none of the solutions are valid, it means that there is no solution,*0626

*because if a solution exists, you will find it with this technique; but it might be buried among extraneous solutions.*0630

*If you find that all of the solutions are extraneous, that means there is no solution to the equation.*0638

*Let's take an example: (x + 1) divided by (x ^{2} + 2x + 15)/(x - 3) = 1/5x.*0645

*Let's just focus on possible extraneous solutions without solving out this whole thing.*0664

*To find the extraneous solutions, you need to think about values*0670

*that would make the denominator 0, because a denominator of 0 is undefined; it is not allowed.*0674

*So, if we ended up with a solution that made the denominator 0, that solution would be extraneous; it would be invalid.*0679

*Factoring out this first denominator, x ^{2} + 2x - 15, is going to give me (x + something) (x - something).*0686

*And factors of 15 are 1 and 15, 3 and 5; and I am looking for factors that will add up to 2.*0700

*5 and -3 would add up to 2; so I am going to go with 5 here and -3 there.*0710

*Recall that the problem would occur if there was a value of x such that this polynomial became 0.*0720

*So, if I had (x + 5) (x - 3) = 0, a value of x that gave me that would be an extraneous solution.*0729

*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.*0736

*So, I am going to set (x + 5) equal to 0 and (x - 3) equal to 0, and solve.*0746

*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.*0754

*OK, so these are some excluded values; and these are based on that first denominator.*0769

*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.*0777

*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.*0789

*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.*0797

*My excluded values are if x equals -5, 3, or 0.*0810

*One way to approach this is to find the excluded values first, when you are working with a rational equation.*0818

*And then, go through and solve, and when you get to your solution, check and make sure it doesn't match an excluded value.*0823

*If it does, you have to throw that solution out.*0831

*If it doesn't match an excluded value, then you are left with a solution that is valid.*0834

*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.*0842

*Recall that a rational inequality is an inequality that contains one or more rational expressions.*0853

*We talked about rational equations containing algebraic fractions, or rational expressions.*0859

*The idea is the same here, except we are dealing with inequalities instead of equalities.*0864

*There are several steps to solve these: the first step is to find the excluded values.*0869

*We talked about excluded values when working with rational equations; and it is the same idea here.*0880

*It is going to be any value of x that is going to make the denominator 0.*0884

*Then, the second step is to solve the related equation.*0889

*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;*0898

*and then, you are going to take the values from steps 1 and 2 and use them to split the number lines into intervals.*0904

*Those intervals may contain parts of the solution set; so then, you are going to use a test value for each interval.*0926

*And that test value is going to help you determine whether or not the interval contains parts of the solution set.*0942

*An example of a rational inequality would be something like x/(x ^{2} - 9) - 3/(x + 4) < 5.*0952

*We will do an example in a second; but pretty much, what you are going to do is find the excluded values*0966

*by finding values that would make the denominator 0, then solve the corresponding equation*0971

*by changing this into an equal sign, and then solving, taking those values (whatever those values are), and then dividing the number line.*0977

*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.*0985

*And then, I would use test points to determine if these intervals contain parts of the solution set.*0994

*I would use the test values and insert them into the original inequality, and see if the inequality held out, based on those values.*1001

*Let's illustrate this right now with an example.*1009

*Given the rational inequality x/(x ^{2} - 9)...actually, let's do a slightly less complex one.*1013

*3/x + 1/(x + 4) ≥ 0*1028

*I am going to start out by finding the excluded values--values that would make the denominator 0.*1041

*For this first one, it is very simple: if x equals 0, then this becomes the denominator 0, and that is not allowable.*1053

*The second one: I get x + 4 = 0; if that were true, this would become an undefined expression.*1061

*Solving for x, x equals -4: if x is -4, add that to 4 and you get 0.*1070

*So, I have two excluded values: x = 0 and x = -4; that is my first step.*1077

*My second step is to solve the related equation, which is 3/x + 1/(x + 4) = 0.*1083

*Recall that we talked about solving rational equations by multiplying all terms by the LCD.*1095

*And these are already factored out; so my LCD is the factor in this denominator, which is unique, times the factor in this denominator,*1105

*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.*1118

*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.*1127

*That is my first term, x(x + 4) times 1/(x + 4) = 0 times x(x + 4).*1137

*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.*1155

*Now, I need to solve this linear equation: this gives me 3x + 12 + x = 0.*1169

*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.*1178

*And that is not an excluded value, so that is a valid solution to this equation.*1193

*So, I have excluded values, and I have a solution to the related equation: x = -3.*1197

*The next step is to use these to divide up the number line into intervals.*1209

*Let's do that down here, out of the way: here x = 0, -1, -2, -3, -4; so, -4: -1, -2, -3...*1216

*Now that I have the number line divided up, I need to determine which of these intervals...*1237

*I have one interval here; I will call this A; it is from right here...this is also dividing it into an interval...*1241

*eliminate these to make the intervals clearer; I have interval B, C, and D; so I have 4 intervals.*1251

*I need to find test points and check those in the original inequality for all 4 of these.*1262

*For the first interval, I am going to use a test point of -5.*1269

*I am working right over here; let's let x equal -5.*1275

*If I go back to this original inequality, let's see if this value over here satisfies it: 3/-5 + 1/(-5 + 4) ≥ 0.*1282

*This gives me -3/5; and this becomes 1/-1, because -5 + 4 is -1.*1295

*So, this gives me -1 3/5, which is not greater than or equal to 0.*1305

*Therefore, this interval does not contain any of the solution set, because my test point did not satisfy this inequality.*1309

*This was for interval A: for interval B, I am going to pick another point, and that is between -3 and -4.*1322

*So, I can't pick an integer; I have to go with a fraction to get something in between there.*1330

*So, I am going to pick -3 1/2; so let's let x equal -3 1/2, which actually equals -7/2.*1336

*This is going to give me 3, divided by -7/2, plus 1, divided by (x + 4), is greater than or equal to 0.*1349

*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.*1364

*Actually, this should be -7/2 right there; and -7/2 + 4 (let's work on this right here)...that would give me...*1384

*I need to convert this, so this would give me 8/2, because 8 divided by 2 would give me 4 back.*1398

*So, that is a common denominator of 2; -7/2 + 8/2 = 1/2.*1404

*That is 3 times -2/7, plus 1/(1/2), is greater than or equal to 0.*1412

*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.*1422

*-6/7 is smaller than 2, so I know that this becomes positive; therefore, this is true.*1439

*When I use x = -3/2, I find that I satisfy the inequality; therefore, B contains part of the solution set.*1448

*So, this value did not satisfy the inequality; this value did satisfy the inequality.*1459

*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,*1467

*because it has to be between 0 and -3; so let's go ahead and use -2.*1479

*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.*1483

*And that becomes -3/2; and this is +1/2, is greater than or equal to 0.*1496

*1/2 and -3/2 gives me -2/2, or -1; is that greater than or equal to 0? No, it is not.*1509

*So, C does not contain the solution set.*1525

*Finally, I am going to test section D, using the test point of 1.*1532

*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.*1536

*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.*1548

*I have determined that the intervals B and D contain the solution set.*1556

*Now, as far as how to write this solution set: let's go up here and write it out.*1561

*You need to be careful that you don't include excluded values.*1572

*The solutions lie between -4 and -3; and even though this says "greater than or equal to,"*1581

*I don't want to include an "equal to" that will include an excluded value.*1589

*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.*1593

*So, x is greater than -4 and less than or equal to -3.*1605

*In addition, x is greater than 0; that is also part of the solution set, which is*1615

*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)...*1630

*see, the -3 can be included in the solution set, because it is not an excluded value.*1638

*It is dividing this up, because it was a solution to this equation, not because it was an excluded value.*1643

*As you can see, this is pretty complicated; it takes a lot of steps.*1649

*Your first step is to find the excluded values; I found those.*1652

*The second step is to solve the related equation; I did that right here, and I got that x equals -3.*1658

*This gave me three values that I used to divide up the number line into four intervals.*1665

*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.*1670

*So, this interval is not part of the solution set; A was right here; B was right here.*1678

*x equals -3/2; I solved that out, and this did satisfy the inequality, so this is part of the solution set.*1685

*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.*1694

*And finally, in D, I let x equal 1; that satisfied the inequality, and therefore, the interval D contains part of the solution set.*1702

*So, the solutions are x > -4 and x ≤ -3, and also x > 0 is part of the solution set.*1712

*Let's get some more practice by doing some examples.*1726

*Going back to rational equations, recall that we need to find the LCD.*1731

*And in order to find the LCD, I am going to factor out these denominators; so this one is factored.*1741

*1 - x is already factored; but if we factor out a -1, it is going to make our work easier.*1753

*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.*1761

*Let's look at this second denominator: I am going to rewrite this, instead of as 1 - x, as -x + 1.*1772

*Now, I can look and see that these two are the same, except the signs of both terms are opposite.*1781

*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.*1786

*-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.*1797

*Now, I can see that I actually have two factors that are the same, plus this -1.*1807

*I could include this 4 as part of the LCD, and then this fraction would end up getting eliminated as well.*1812

*But I am not really worried about this fraction, because it is a constant; I can work with the number 1/4.*1820

*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.*1824

*I am actually going to not use that 4, and I am just going to deal with the constant later on.*1833

*But we still need to multiply this term by the LCD that we find here, as well.*1837

*OK, so the LCD is going to be the product of (x - 1), because that is a factor, and -1.*1844

*So, I am going to multiply each term in this equation by this LCD.*1854

*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.*1870

*I just factored this into this by factoring out the -1.*1880

*OK, let's go ahead and start canceling common factors.*1894

*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.*1897

*So, this gives me -4: so -3 - 4 equals -1, times (x - 1), all divided by 4.*1914

*I am not worried about this, because the denominator does not have a variable in it, so it is not a rational expression.*1929

*I have gotten rid of all of my rational expressions: remember, rational expressions are algebraic fractions,*1934

*but we are talking about fractions where there is a variable in the denominator.*1938

*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.*1943

*That will cancel this -4 out, and that gives me -4 times -7, which is 28, equals x - 1.*1959

*I am going to add a 1 to both sides to get 29 = x.*1968

*Now, I can't forget about my excluded values: these are going to be values that make the denominator 0.*1974

*And for this first one, I have (x - 1); if that equals 0, then this will be undefined.*1983

*So, that would occur in cases where x equals 1; so x = 1 is an excluded value.*1991

*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.*1997

*x = 1 is the excluded value for this one and for this one.*2009

*So, since this solution I got, x = 29, is not an excluded value, then it is a valid solution.*2013

*If I came up with the solution x = 1, that would have been an extraneous solution that I would have had to throw out.*2021

*This example is another rational equation that we are again going to solve by finding the LCD.*2032

*So, let's go ahead and just factor the denominators right here.*2040

*This is x plus a factor, times x minus a factor; factors of 10 are 1 and 10, 2 and 5.*2044

*And I need those to add up to 3x; so 5 - 2 would give me a 3.*2054

*Therefore, the correct factorization would be (x + 5) (x - 2), plus...I am going to leave some room to multiply these by the LCD.*2062

*So, 2/(x - 2) = 19/(x + 5): I have this factored out right here.*2078

*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.*2091

*So, I am going to multiply each term by that LCD, (x + 5) (x - 2),*2107

*plus (x + 5) (x - 2) times this next term, times (x + 5) (x - 2) times this third term.*2118

*Go ahead and cancel out common factors: here, both factors are common, leaving a 3 behind.*2135

*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).*2143

*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.*2158

*Combining the constants on the left, 2x + 13 = 19x - 38.*2176

*I am going to subtract the 2x, first, from each side: that is going to give me 13 = 17x - 38.*2185

*And then, I am going to go ahead and add a 38 to both sides to give me 51 = 17x.*2197

*Divide both sides by 17; this is going to give me 51/17 = x; therefore, x = 3.*2205

*Now, before I say that this is the actual solution, I need to look for excluded values.*2215

*And it is easy to do that, because I have already factored these out.*2219

*Excluded values: this is something you can do right at the beginning, before you start working, or right when you finish.*2223

*But you have to make sure that you check the solutions, each time, for these.*2230

*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.*2234

*So, using the zero product property, if (x + 5) equals 0, this whole thing will equal 0.*2245

*Or if (x - 2) equals 0, this whole denominator will equal 0, when you multiply 0 times the other factor.*2254

*So, x = -5 and x = 2 are excluded values.*2261

*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.*2269

*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.*2277

*Excluded values for all 3 (I have covered all 3) are these two.*2284

*I look over here, and my solution, x = 3, is not an excluded value; therefore, it is a valid solution.*2289

*OK, again, we have a rational equation we need to solve.*2299

*And the first step is to find the LCD by factoring out the denominators.*2302

*So, this first denominator (we will find the LCD)...I just have x + 1; that is already factored.*2308

*x ^{2} - 1 is the difference of two squares, so that is (x + 1) (x - 1).*2318

*Therefore, the LCD...I have this (x + 1) factor here and here, and I have an x - 1.*2325

*So, I am going to solve this by multiplying each term by (x + 1) (x - 1)...times x, equals (x + 1) (x - 1)*2335

*times (x ^{2} + x + 2)/(x + 1), minus (x + 1) (x - 1) times (x^{2} - 5), divided by...*2351

*I am going to rewrite this in the factored form, so it is more obvious what cancels and what doesn't.*2368

*All right, so over here, I end up with just x, times (x + 1) (x - 1).*2376

*Here, the (x + 1)s cancel out; that is going to leave me with (x - 1) times (x ^{2} + x + 2),*2386

*minus...the (x + 1)s cancel; the (x - 1)s cancel; so minus (x ^{2} - 5).*2397

*Multiplying this out, recall that (x + 1) (x - 1) is (x ^{2} - 1), so this gives me x times (x^{2} - 1).*2406

*equals...x times x ^{2} is x^{3}; x times x is x^{2}; x times 2 is 2x.*2416

*Multiplying -1 times each of these terms: -x ^{2} - x...-1 times 2 is -2.*2425

*A negative times a positive gives me -x ^{2}; a negative times a negative gives me + 5.*2435

*Now, to take care of this: this is x ^{3} - x.*2441

*All right, first let's take care of the x ^{3}; I am going to subtract x^{3} from both sides to move this from the right to the left.*2457

*So, what happens is: I do x ^{3} - x^{3}; this drops out.*2464

*I have to do the same thing to the left: x ^{3} - x^{3}--that drops out.*2470

*So, that took care of the x ^{3}; this looked worse than it actually was.*2475

*It is leaving me with -x =...well, let's see if I can do any simplifying here on the right.*2480

*I have x ^{2} and -x^{2}, so those cancel out.*2485

*That leaves me with 2x...so this is gone; this is gone; this is gone...minus x, minus 2, minus x ^{2}, plus 5.*2501

*I still have some work to do; but I am going to see that I can combine some of this still,*2524

*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 x ^{2}.*2532

*Now, what I am going to end up with here is a quadratic equation that I need to solve.*2553

*Let's finish doing the simplification, and then go ahead and solve that.*2558

*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 - x ^{2}.*2561

*I want to write this in a more standard form, where the x ^{2} is going to be positive.*2572

*So, let's go ahead and flip the sides, which will give me x ^{2} - 2x - 3 = 0.*2577

*All I did is added an x ^{2} to both sides, subtracted a 3 from both sides, and subtracted a 2x from both sides, in order to flip this around.*2586

*OK, now that I have just a quadratic equation left, all I need to do is solve it.*2598

*And I can use factoring: this is negative, so I have (x + a factor), times (x - a factor), equals 0.*2602

*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.*2614

*And this does work; you get x ^{2}...the outer term is -3x...plus x; that gives a -2x for the middle term; 1 times -3 is -3.*2624

*That is factored correctly; now I just need to use the zero product property to solve.*2635

*x + 1 = 0; therefore, x = -1; the other solution is that x - 3 = 0; so x = 3.*2640

*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.*2650

*Possible solutions are x = -1 and x = 3.*2658

*Now, let's look for excluded values: excluded values are going to be any values that make either of the denominators 0.*2665

*Here, I have x ^{2} - 1; that factored into (x + 1) (x - 1).*2679

*So, if (x + 1) (x - 1) equals 0, the denominator becomes 0.*2685

*So, any values of x that result in this being 0 are excluded.*2690

*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.*2696

*Therefore, for this first one, if x equals -1, this whole thing will become 0,*2711

*because -1 times -1 is 1, minus 1--this would give you a 0.*2719

*The other excluded value is x = 1, because 1 ^{2} is 1, minus 1 would give you 0.*2724

*So, I have these two excluded values: looking over here, I already took care of that, because this is a factor, right here.*2729

*So, I already said that, if x + 1 equals 0, x equals -1, and that is excluded; so that one is already covered.*2737

*Now, I need to compare my solutions to my possible solutions, to these.*2743

*And I see that I only have one valid solution, because this is actually not a valid solution.*2749

*It is an excluded value; I can't include that as part of the solution--it is an extraneous solution.*2758

*x = 3 is the only valid solution.*2764

*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.*2768

*And then, the hardest work was actually just multiplying all of this out, keeping track of all the signs,*2777

*and then getting down to where we had a quadratic equation that we solve by factoring to get two possible solutions.*2783

*The final step was to find the excluded values, based on setting the denominators equal to 0 and then solving for x,*2788

*and then comparing my possible solutions against these excluded values, thus eliminating x = -1 from the solution set,*2796

*and being left with one solution, which is x = 3.*2804

*In Example 4, we are going to be working with a rational inequality.*2809

*Recall that, for rational inequalities, we are going to find the excluded values.*2814

*We are going to solve the related equation; and then we are going to use those values to divide the number line into intervals,*2819

*and then test each interval, using a test value to determine where the solution set is.*2824

*Let's first find excluded values: these are going to be values for which the denominator becomes 0.*2832

*If 2x equals 0, that would occur when x equals 0/2, or x equals 0.*2846

*The same thing is going to occur here with 4x--that if x equals 0, it will be an excluded value.*2853

*So, the excluded value is x = 0.*2858

*Now, I need to solve the related equation, 1/2x + 5/4x - 1 = 0.*2862

*And the least common denominator of 2x and 4x...I could factor this out to 2 times 2.*2874

*So, the LCD is going to be...there is a 2 present once here, but twice here; and remember, I take each unique factor*2887

*to the highest power that it is present, so I am going to say 2 ^{2}.*2898

*There is an x present once here and once here, so it is 2 ^{2} times x equals 4x.*2902

*So, the LCD is 4x; so multiplying here, 4x(1/2x) + 4x(5/4x) - 1(4x) = 0(4x).*2907

*Canceling out common factors, this becomes a 2; 2 cancels out; the x's cancel out.*2928

*2 times 1 is 2, plus...the 4x's cancel, leaving behind 5; this is -4x = 0.*2935

*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.*2949

*So, that is my excluded value; and my solution to the related equation is x = 1 3/4.*2968

*Let's go down here, and I am going to use this value and this value to divide up my number line.*2974

*I have 0 here, and I have 1 3/4 here.*2984

*So, I end up with three intervals: A, B, and C; and I need to test a point within each of these intervals.*2992

*And if that value satisfies the inequality, I know that this interval is at least part of the solution set.*3003

*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.*3010

*That is going to give me 1 over 2(-1), plus 5 divided by 4(-1), minus 1, is greater than 0.*3022

*And let's see if this holds up: this gives me -1/2, minus 5/4, minus 1, is greater than 0.*3031

*This is the same as saying -2/4 (just converting to a common denominator) - 5/4 - 1 > 0.*3040

*I actually don't even need to go farther; in fact, I even kind of just looked at this and said,*3049

*"I have a bunch of negatives; those are not going to be greater than 0 when I combine them."*3053

*So, no: this is not valid; therefore, that interval does not contain the solution set or part of the solution set.*3057

*For B, I have values between 0 and 1 3/4, so I can use 1 as a test point, x = 1.*3069

*This is going to give me 1 over 2(1), plus 5 divided by 4(1), minus 1, is greater than 0,*3077

*which is 1/2 + 5/4 - 1 > 0; I can just convert this to 2/4 + 5/4 - 1 > 0.*3086

*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.*3101

*So, this is going to give me 3/4 > 0.*3116

*Even if I didn't figure this whole thing out, as soon as I saw that this is a positive number larger than 1,*3119

*I know that, when I subtract 1 from it, I will get something positive; so it is going to be greater than 0.*3125

*So, this one is valid; therefore, yes for B: this interval does contain at least part of the solution set.*3129

*For C, I am going to go right here, and I am going to use 2 as my test point: x = 2.*3138

*This is going to give me 1 over 2(2), plus 5 over 4(2), minus 1, is greater than 0.*3148

*That is 1/4 + 5/8 - 1 > 0.*3156

*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.*3163

*That is 7/8 - 1 > 0; well, 7/8 - 1 is going to leave -1/8 > 0.*3172

*And that is not true, so C is not part of the solution set.*3183

*Therefore, the solution for this...the possible solution...I am going to say "possible,"*3188

*because we have to look back at excluded values...is x is greater than 0, but it is less than 1 3/4.*3196

*Now, let's look back up at excluded values: I cannot let x equal 0--that is an excluded value.*3209

*But I am OK here, because x is greater than 0; so I am covered, and this is my actual solution--this is valid.*3223

*And just looking back to review: find your excluded values (x = 0 is an excluded value), then solve the related rational equation.*3234

*I did that by multiplying by the LCD, multiplying each term, getting rid of those fractions,*3247

*then finding that x = 1 3/4 was the solution, and it wasn't an excluded value (so it was a valid solution).*3253

*I used those two values, 0 and 1 3/4, to divide the number line into intervals.*3262

*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.*3268

*I tested B, using x = 1, and found the inequality did hold up, so B contains part of my solution set.*3276

*C--I tested: that value did not satisfy the inequality, so that is not part of the solution set.*3284

*Therefore, the solution set is that x is greater than 0 and less than 1 3/4.*3289

*But remember: if you are working with greater than or equal to in the original, and you are thinking,*3293

*"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.*3300

*That concludes this lesson of Educator.com on rational equations and inequalities.*3307

*I will see you next lesson!*3314

0 answers

Post by Samvel Karapetyan on July 8, 2012

Great work you're doing sweet doctor.

1 answer

Last reply by: Dr Carleen Eaton

Mon Feb 6, 2012 11:18 PM

Post by Edmund Mercado on February 5, 2012

Dr. Eaton:

At 22:25, I think you meant to write -3.5 instead of -3/2 for -7/2.