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INSTRUCTORSCarleen EatonGrant Fraser
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Lecture Comments (3)

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.

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

Solve:[5/3] − [1/3] = [(n + 1)/(3n + 6)]
  • 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
n = [( − 21)/9] = − [7/3]
Solve:[1/(2x2 − 14x + 12)] + [5/(2x − 2)] = [1/(x − 6)]
  • List the prime factors of 2x2 − 14x + 12, 2x − 2 and x − 6 to find the LCD. Factoring may be required.
  • 2x2 − 14x + 12 =
  • 2x − 2 =
  • x − 6 =
  • 2x2 − 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/(2x2 − 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
x = 9
Solve:[x/(x − 6)] − [1/(3x − 18)] = [1/3]
  • 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
x = − [5/2]
Solve:[1/(2x − 5)] − [1/(x − 2)] = [(x + 5)/(2x2 − 9x + 10)]
  • List the prime factors of 2x − 5;x − 2 and 2x2 − 9x + 10 to find the LCD. Factoring may be required.
  • 2x − 5 =
  • x − 2 =
  • 2x2 − 9x + 10 =
  • 2x − 5 = (2x − 5)
  • x − 2 = (x − 2)
  • 2x2 − 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)/(2x2 − 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
x = − 1
Solve:[3/x] + [1/(x2 + x)] = [1/(x + 1)]
  • List the prime factors of x, x2 + x and x + 1 to find the LCD. Factoring may be required.
  • x =
  • x2 + x =
  • x + 1 =
  • x = x
  • x2 + 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/(x2 + 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
x = − 2
Solve:[6/x] − [1/(x2 − 6x)] = [1/(x − 6)]
  • List the prime factors of x, x2 − 6x and x − 6 to find the LCD. Factoring may be required.
  • x =
  • x2 − 6x =
  • x − 6 =
  • x = x
  • x2 − 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/(x2 − 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
x = [37/5]
Solve:[(x − 6)/(x − 1)] + [4/(x2 − 1)] = 1
  • List the prime factors of x − 1, x2 − 1 to find the LCD. Factoring may be required.
  • x − 1 =
  • x2 − 1 =
  • x − 1 = (x − 1)
  • x2 − 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/(x2 − 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))]
  • [(x2 − 5x − 2)/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
  • Subtract the right side of the eqution
  • [(x2 − 5x − 2)/((x − 1)(x + 1))] − [((x − 1)(x + 1))/((x − 1)(x + 1))] = 0
  • Subtract
  • [(x2 − 5x − 2 − (x2 − 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]
Solve: [x/(x − 6)] − [1/(3x − 18)] < [1/3]
  • 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
− [5/2] < x < 6
Solve: [(x − 6)/(x − 1)] + [4/(x2 − 1)] ≤ 1
  • Identify excluded values
  • x − 1 = 0
  • x = 1
  • x2 − 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/(x2 − 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))]
  • [(x2 − 5x − 2)/((x − 1)(x + 1))] = [((x − 1)(x + 1))/((x − 1)(x + 1))]
  • Subtract the right side of the eqution
  • [(x2 − 5x − 2)/((x − 1)(x + 1))] − [((x − 1)(x + 1))/((x − 1)(x + 1))] = 0
  • Subtract
  • [(x2 − 5x − 2 − (x2 − 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/(x2 − 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/(x2 − 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/(x2 − 1)] ≤ 1
    [(0 − 6)/(0 − 1)] + [4/(02 − 1)] ≤ 1
    6 − 4 ≤ 1
    Not True
  • Test: x=2
    [(x − 6)/(x − 1)] + [4/(x2 − 1)] ≤ 1
    [(2 − 6)/(2 − 1)] + [4/(22 − 1)] ≤ 1
    − 4 + [4/3] ≤ 1
    True
  • Therefore, the solution is
− 1 < x ≤ − [1/5] and x > 1

*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 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

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: (x2 - 3x + 9)/(x2 - 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 x2, for example, here, then I would have written this as 2x2(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 92 - 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/(x2 + 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

x2 + 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 x2 - 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 equations0580

(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 (x2 + 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 values0670

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, x2 + 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/(x2 - 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 values0966

by finding values that would make the denominator 0, then solve the corresponding equation0971

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/(x2 - 9)...actually, let's do a slightly less complex one.1013

3/x + 1/(x + 4) ≥ 01028

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 is1615

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

x2 - 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 (x2 + x + 2)/(x + 1), minus (x + 1) (x - 1) times (x2 - 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 (x2 + x + 2),2386

minus...the (x + 1)s cancel; the (x - 1)s cancel; so minus (x2 - 5).2397

Multiplying this out, recall that (x + 1) (x - 1) is (x2 - 1), so this gives me x times (x2 - 1).2406

equals...x times x2 is x3; x times x is x2; x times 2 is 2x.2416

Multiplying -1 times each of these terms: -x2 - x...-1 times 2 is -2.2425

A negative times a positive gives me -x2; a negative times a negative gives me + 5.2435

Now, to take care of this: this is x3 - x.2441

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.2457

So, what happens is: I do x3 - x3; this drops out.2464

I have to do the same thing to the left: x3 - x3--that drops out.2470

So, that took care of the x3; 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 x2 and -x2, 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 x2, 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 x2.2532

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 - x2.2561

I want to write this in a more standard form, where the x2 is going to be positive.2572

So, let's go ahead and flip the sides, which will give me x2 - 2x - 3 = 0.2577

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.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 x2...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 x2 - 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 12 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 factor2887

to the highest power that it is present, so I am going to say 22.2898

There is an x present once here and once here, so it is 22 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