6. Solving Rational Equations and Inequalities
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Chapter 4
6.
Solving Rational Equations and Inequalities
Rational equations and inequalities play a pivotal role in mathematical studies. This lesson delves into methods for solving equations that contain one or more rational expressions. While the techniques for solving these equations have been previously discussed, the focus here is on inequalities containing rational expressions. Real-life scenarios, such as Dominika's winning record in basketball games and her chemistry experiments, are used to illustrate the application of these mathematical concepts. Additionally, the lesson touches upon strategies like using the Cross Products Property and finding the Least Common Denominator. It is essential to verify solutions to ensure they are not extraneous, meaning they do not fit within the domain of the given problem.
Show less Show more expand_more| Student Learning Objectives: |
|---|
|
| | 11 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Rational Equations and Inequalities
Slide of 11
This lesson focuses on solving equations that contain one or more rational expressions. Since methods for solving this type of equations has already been covered, this lesson will focus on inequalities that contain rational expressions.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Dominika's Winning Percentage in Basketball
Heichi and Dominika like to play basketball. About two months ago, they decided to keep track of how many games they each win. Until now, Dominika has won 24 out of the 40 games against Heichi.
a
How many games would Dominika have to win in a row to have a 75 % winning record?
b
How many games would Dominika have to win in a row to have a 90 % winning record?
c
Is Dominika able to reach a 100 % winning record? Explain why or why not.
d
Suppose that after reaching a winning record of 90 %, Dominika had a losing streak. How many games in a row would Dominika have to lose to drop down to a winning record below 50 % again?
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Discussion
Rational Equations and Finding Their Solutions
A rational equation is an equation that contains at least one rational expression. 1/x + x/2 = x+4/2x If a rational equation is expressed as a proportion, it can be solved by using the Cross Products Property. Otherwise, to eliminate the fractions, each side of the equation is multiplied by the least common denominators of the expressions.
| Rational Equation | Method |
|---|---|
| x/2x+7 = 2/x-1 | Cross Products Property |
| x/x-5+x/x+5 = 2/x^2-25 | LCD |
Method
Solving a Rational Equation by Using the Cross Products Property
Rational equations can be solved by using different methods. The main idea is to move the variables out of the denominators. Furthermore, since the domain of rational functions is restricted, some solutions might be extraneous. Therefore, it is necessary to verify the solutions in the original equation. Consider this rational equation as an example. - 1/x-3 = - x+4/x^2-2 The equation is basically a proportion. Therefore, it can be solved by using the Cross Products Property.
1
Use the Cross Products Property
expand_more To get the variable terms out of the denominators, the Cross Product Property will be used.
- 1/x-3 = - x+4/x^2-2 ⇓ -1* (x^2-2) = (x-3)(- x+4)
As shown, the property eliminates the rational expressions in the equation. Now the equation can be solved for x.
2
Solve the Resulting Equation
expand_more To solve the resulting equation, use the Distributive Property and combine like terms.
-1* (x^2-2) =(x-3)(- x+4)
Distr
Distribute - 1
- x^2+2 = (x-3)(- x+4)
- x^2+2 = - x^2+7x-12
▼
Solve for x
AddEqn
LHS+x^2=RHS+x^2
2 = 7x-12
AddEqn
LHS+12=RHS+12
14=7x
DivEqn
.LHS /7.=.RHS /7.
2 =x
RearrangeEqn
Rearrange equation
x = 2
The value that satisfies the equation is 2. Next, check if it also satisfies the original equation.
3
Check for Extraneous Solutions
expand_more The solution found in the previous step might be extraneous. It must be verified in the original equation.
- 1/x-3 = - x+4/x^2-2
Substitute
x= 2
- 1/2-3 ? = - 2+4/2^2-2
1 ? = - 2+4/2^2-2
▼
Evaluate right-hand side
CalcPow
Calculate power
1 ? = - 2+4/4-2
AddSubTerms
Add and subtract terms
1 ? = 2/2
CalcQuot
Calculate quotient
1 = 1 ✓
Substiting x=2 in the original equation produced a true statement. Therefore, it is a solution to the original equation.
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Example
Creating a 45 % Acid Solution
In her chemistry lab, Dominika adds some 60 % acid solution to 15 milliliters of a solution with 15 % acid.
How much of the 60 % acid solution should she add to create a solution that is 45 % acid?
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Hint
The percentage of acid in the final solution must equal the total amount of acid divided by the total amount of solution.
Solution
Let x be the amount of 60 % acid solution to be added to 15 milliliters of a 15 % acid solution. The amount of each solution in terms of x can be organized in a table.
| Original | Added | New | |
|---|---|---|---|
| Amount of Acid | 0.15(15) | 0.6(x) | 0.15(15)+0.6(x) |
| Total Solution | 15 | x | 15+x |
The percentage of acid in the final solution must equal the amount of acid divided by the total solution. Percent=Amount of Acid/Total Solution The objective is to create a solution that is 45 % acid, so place that on the left side of the equation. Then substitute the known values into the rest of the formula. 45/100=0.15(15)+0.6(x)/15+x This proportion is also a rational expression. To solve the equation for x, the Cross Products Property can be used.
45/100=0.15(15)+0.6(x)/15+x
Multiply
Multiply
45/100=2.25+0.6x/15+x
CrossMult
Cross multiply
45(15+x)=100(2.25+0.6x)
x = 30
The solution might be extraneous, so it must be verified in the original equation.
Substituting x=30 in the original equation produced a true statement, so it is the solution to the equation. Dominika should add 30 milliliters of the 60 % acid solution.
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Discussion
Solving a Rational Equation by Using the Least Common Denominator
To solve a rational equation, the first step is to eliminate all the fractions in the equation. This can be done by multiplying each side of the equation by the least common denominator of the rational expressions. Any time an equation is multiplied by an algebraic expression, an extraneous solution might be introduced, so it will also be necessary to verify the solutions. 5/x+2 - 4/x+5 = 16/x^2+7x+10 Consider solving the this equation as an example.
1
Find the Least Common Denominator
expand_more To find the least common denominator (LCD) of the rational expressions in the equation, their factored form must be found. 5/x+2 - 4/x+5 = 16/x^2+7x+10
The denominator of the rational expression on the left-hand side cannot be factored further. Factor the denominator of the other expression.
x^2+7x+10
(x+2)(x+5)
Since the LCD is the product of the factors with the highest power appearing in any denominator, the LCD is (x+2)(x+5).
| Rational Expression | LCD |
|---|---|
| 5/x+2 | (x+2)(x+5) |
| 4/x+5 | |
| 16/(x+2)(x+5) |
2
Multiply the Equation by the Least Common Denominator
expand_more Multiply each side of the equation by the LCD found in the previous step. This will reduce each term and get rid of the denominators.
5/x+2 - 4/x+5 = 16/x^2+7x+10
MultEqn
LHS * (x+2)(x+5)=RHS* (x+2)(x+5)
(x+2)(x+5)(5/x+2 - 4/x+5 )= (x+2)(x+5)(16/x^2+7x+10)
▼
Simplify
Distr
Distribute (x+2)(x+5)
5(x+2)(x+5)/x+2 - 4(x+2)(x+5)/x+5 = 16(x+2)(x+5)/x^2+7x+10
CancelCommonFac
Cancel out common factors
5(x+2)(x+5)/x+2 - 4(x+2)(x+5)/x+5 = 16(x+2)(x+5)/x^2+7x+10
SimpQuot
Simplify quotient
5(x+5)-4(x+2) = 16
3
Solve the Resulting Equation
expand_more Now the resulting equation can be solved by using the Distributive Property and inverse operations.
5(x+5)-4(x+2) = 16
Distr
Distribute 5
5x+25-4(x+2) = 16
Distr
Distribute - 4
5x+25-4x-8 = 16
SubTerms
Subtract terms
x + 17 = 1 6
SubEqn
LHS-17=RHS-17
x = - 1
4
Check for Extraneous Solutions
expand_more Finally, check if the solution found in Step 3 is an extraneous solution or not. To do so, substitute it in the original equation and simplify.
5/x+2 - 4/x+5 = 16/x^2+7x+10
Substitute
x= - 1
5/- 1+2 - 4/- 1+5 ? = 16/( - 1)^2+7( - 1)+10
4 ? = 16/(- 1)^2+7(- 1)+10
▼
Evaluate right-hand side
CalcPow
Calculate power
4 ? = 16/1+7(- 1)+10
MultPosNeg
a(- b)=- a * b
4 ? = 16/1-7+10
AddSubTerms
Add and subtract terms
4 ? = 16/4
CalcQuot
Calculate quotient
4 = 4 ✓
The value made a true statement. Therefore, it is a solution to the equation.
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Example
Canoe Trip
Dominika and Heichi are taking a canoe trip. They are going up the river for 1 kilometer and then returning to their starting point. The river current flows at 3 kilometers per hour. The total trip time will be 1 hour and 30 minutes.
Assuming that they paddled at a constant rate throughout the trip, find the speed at which Dominika and Heichi are paddling. Round the answer to the two decimal places.
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Hint
Let x represent the speed, in kilometers per hour, that the canoe would travel with no current. Write two rational expressions for the time it takes to go and return in terms of x.
Solution
The two students are on the river for 1 hour and 30 minutes, or 1.5 hours. This round trip time represents the time it takes to go upriver and return. Recall the formula that relates distance to time, d=rt. This formula can be maniuplated to represent distance in terms of time.
d = rt ⇔ t = d/r
Let x represent the speed, in kilometers per hour, that the canoe would travel with no current. When Dominika and Heichi are traveling with the current, their speed is x+3 km/h, and when they travel against the current, their speed is x-3 km/h. The trip takes 1x+3 hours for the outbound portion and 1x-3 hours for the return portion.
| Going | Returning | |
|---|---|---|
| Distance (km) | 1 | 1 |
| Speed (km/h) | x+3 | x-3 |
| Time (h) | 1/x+3 | 1/x-3 |
The sum of the portions is equal to the total round trip time. 1/x+3 + 1/x-3 = 1.5 The least common denominator of the rational expressions is (x+3)(x-3). To eliminate the rational denominators, multiply each side of the equation by the LCD.
1/x+3 + 1/x-3 = 1.5
MultEqn
LHS * (x+3)(x-3)=RHS* (x+3)(x-3)
(x+3)(x-3) (1/x+3 + 1/x-3 ) = (x+3)(x-3)1.5
▼
Simplify left-hand side
Distr
Distribute (x+3)(x-3)
(x+3)(x-3)/x+3 + (x+3)(x-3)/x-3 = (x+3)(x-3)1.5
CancelCommonFac
Cancel out common factors
(x+3)(x-3)/x+3 + (x+3)(x-3)/x-3 = (x+3)(x-3)1.5
SimpQuot
Simplify quotient
x-3+x+3 = (x+3)(x-3)1.5
AddTerms
Add terms
2x = (x+3)(x-3)1.5
▼
Simplify right-hand side
2x = 1.5x^2-13.5
SubEqn
LHS-2x=RHS-2x
0 = 1.5x^2-2x-13.5
RearrangeEqn
Rearrange equation
1.5x^2-2x-13.5 = 0
To get rid of decimals in the equation, both sides of the equation can be multiplied by 2. The equation can then be solved by using the Quadratic Formula.
1.5x^2-2x-13.5 = 0
MultEqn
LHS * 2=RHS* 2
3x^2-4x-27 = 0
▼
Solve using the quadratic formula
UseQuadForm
Use the Quadratic Formula: a = 3, b= - 4, c= -27
x_(1,2)=-( - 4)±sqrt(( - 4)^2 - 4( 3)( -27))/2( 3)
NegNeg
- (- a)=a
x_(1,2)=4±sqrt(( - 4)^2 - 4( 3)( -27))/2( 3)
CalcPowProd
Calculate power and product
x_(1,2)=4±sqrt(16 + 324)/6
AddTerms
Add terms
x_(1,2)=4±sqrt(340)/6
SplitIntoFactors
Split into factors
x_(1,2)=4±sqrt(4 * 85)/6
SqrtProd
sqrt(a* b)=sqrt(a)*sqrt(b)
x_(1,2) = 4 ± sqrt(4) * sqrt(85)/6
CalcRoot
Calculate root
x_(1,2) = 4 ± 2 * sqrt(85)/6
FactorOut
Factor out 2
x_(1,2) = 2(2 ± sqrt(85))/6
SimpQuot
Simplify quotient
x = 2 ± sqrt(85)/3
The solutions for this equation are x= 2 ± sqrt(85)3. Separate them into the positive and negative cases and round the answers to two decimal places.
| x=2±sqrt(85)/3 | |
|---|---|
| x_1=2+sqrt(85)/3 | x_2=2-sqrt(85)/3 |
| x_1=2/3+sqrt(85)/3 | x_2=2/3-sqrt(85)/3 |
| x_1 ≈ 3.74 | x_2≈ - 2.41 |
The second solution does not make sense in the given context as speed cannot be negative. The first solution must be verified in the original equation to determine if it is extraneous or not. If it is a valid solution, substituting x=3.74 should equal about 1.5.
1/x+3 + 1/x-3 ? = 1.5
Substitute
x= 3.74
1/3.74+3 + 1/3.74-3 ? = 1.5
▼
Evaluate left-hand side
AddTerms
Add terms
1/6.74 + 1/3.74-3 ? = 1.5
SubTerm
Subtract term
1/6.74 + 1/0.74 ? = 1.5
ExpandFrac
a/b=a * 0.74/b * 0.74
0.74/4.9876 + 1/0.74 ? = 1.5
ExpandFrac
a/b=a * 6.74/b * 6.74
0.74/4.9876 + 6.74/4.9876? = 1.5
AddFrac
Add fractions
7.48/4.9876 ? = 1.5
CalcQuot
Calculate quotient
1.499719 ... ? = 1.5
RoundInt
Round to nearest integer
1.5 = 1.5 ✓
Substituting the solution into the original equation resulted in an identity, so this is a solution to the original equation. Therefore, the speed at which the students can row their canoe is about 3.74 kilometers per hour.
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Example
Identifying an Extraneous Solution
Heichi is trying to solve the following rational equation. 2/x^2-1 - 1/x-1 = 1/x+1 However, he cannot be sure about his solution, so he shares his thoughts with Dominika.
|
If I clear the denominators I find that the only solution is x=1, but when I substitute x=1 into the equation, it does not make any sense. |
Is Heichi correct? Explain.
Answer
Yes, see solution.
Hint
Solve the given rational equation. Is the solution in the domain of the rational expressions on both sides?
Solution
To check if Heichi's claim is correct or not, the given equation will be solved by using the least common denominator. Notice that x^2−1=(x+1)(x−1), so x^2−1 is the common denominator for the three rational expressions. Multiply the equation by this common denominator and solve for x.
2/x^2-1 - 1/x-1 = 1/x+1
MultEqn
LHS * (x-1)(x+1)=RHS* (x-1)(x+1)
(x-1)(x+1) (2/x^2-1 - 1/x-1) = (x-1)(x+1)(1/x+1 )
▼
Solve for x
Distr
Distribute (x-1)(x+1)
2(x-1)(x+1)/x^2-1 - (x-1)(x+1)/x-1 = (x-1)(x+1)(1/x+1 )
MoveLeftFacToNumOne
a* 1/b= a/b
2(x-1)(x+1)/x^2-1 - (x-1)(x+1)/x-1 = (x-1)(x+1)/x+1
FacDiffSquares
a^2-b^2=(a+b)(a-b)
2(x-1)(x+1)/(x-1)(x+1) - (x-1)(x+1)/x-1 = (x-1)(x+1)/x+1
CancelCommonFac
Cancel out common factors
2(x-1)(x+1)/(x-1)(x+1) - (x-1)(x+1)/x-1 = (x-1)(x+1)/x+1
SimpQuot
Simplify quotient
2-(x+1)=x-1
Distr
Distribute - 1
2-x-1 = x-1
AddEqn
LHS+1=RHS+1
2-x = x
AddEqn
LHS+x=RHS+x
2 = 2x
DivEqn
.LHS /2.=.RHS /2.
1 = x
RearrangeEqn
Rearrange equation
x =1
Heichi is correct that when the denominators are eliminated, the solution to the resulting equation is 1. However, the rest of his statement is also correct because the denominators of the expressions on the left hand side are 0 when x=1. 2/1^2-1_0 - 1/1- 1_0 = 1/x+1 Therefore, the value Heichi found is not in the domain of the rational expressions on the left-hand side, which makes it an extraneous solution.
Extra
Why x=1 is an extraneous solution?An alternative explanation for why x=1 are extraneous solutions can be explained as follows. In the example, both sides of the equation are multiplied by x^2-1 to create an equivalent equation.
Multiplying both sides by x^2-1 is legitimate unless x=1 or x=- 1. When x=1 or x=- 1, this is equivalent to multiplying both sides by 0. Therefore, the operation only produces an equivalent equation if those two values are excluded. Excluded Values x = -1 and x=1 After the equation was solved for x, it was found that x=1 is a solution. However, since x=1 is excluded, it is an extraneous solution.
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Example
Finding the Time Spent Painting the Canoe
Dominika and Heichi want to paint their canoe before their next trip up the river.
a
Dominika thinks that if she works alone, it would take her 3 times as long as it would take Heichi to paint the entire canoe by himself. However, by working together, they can complete the job in 3 hours. How long would it take Heichi to complete the job?
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b
Heichi thinks that if he works alone, it would take him 6 hours less than it would take Dominika to paint the entire canoe by herself. If they worked together, they could complete the job in 4 hours. How long would it take Dominika to paint the canoe if Heichi were not there to help?
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Hint
a
Let t be the time it takes Heichi to complete the work. The amount of work done W is the product of the rate of work r and the time spent working t.
b
The amount of work done W is the product of the rate of work r and the time spent working t.
Solution
a
Let t be the time it takes Heichi to complete the work alone. Then, the time it would take Dominika to complete the work by herself would be 3t. The work in this case is painting 1 canoe, or 1. Now an expression to represent each person’s rate can be written using the formula r= Wt.
c|c Dominika's Rate & Heichi's Rate 1/3t & 1/t Dominika thinks that it would take them 3 hours to paint the canoe if they work together. If their combined hourly rate, the sum of the individual ratios, is multiplied by 3, it should be 1, which is the number of canoes they could paint in 3 hours. 3 (1/3t + 1/t ) =1 Now solve the rational equation for x.
3 (1/3t + 1/t ) =1
▼
Solve for t
ExpandFrac
a/b=a * 3/b * 3
3 (1/3t + 3/3t ) =1
AddFrac
Add fractions
3 (4/3t ) =1
Multiply
Multiply
12/3t = 1
MultEqn
LHS * 3t=RHS* 3t
12 = 3t
DivEqn
.LHS /3.=.RHS /3.
4 = t
RearrangeEqn
Rearrange equation
t = 4
Now check the solution in the original equation to confirm whether it is an extraneous solution.
3 (1/3t + 1/t ) =1
Substitute
t= 4
3 (1/3( 4) + 1/4 ) ? = 1
▼
Evaluate left-hand side
Multiply
Multiply
3 (1/12 + 1/4 ) ? = 1
ExpandFrac
a/b=a * 3/b * 3
3 (1/12 + 3/12 ) ? = 1
AddFrac
Add fractions
3 (4/12 ) ? = 1
Multiply
Multiply
12/12 ? = 1
CalcQuot
Calculate quotient
1 = 1 ✓
Checking the solution resulted in an identity, so t=4 is a solution. According to Dominika, Heichi would take 4 hours to paint the canoe by himself. Dominika’s time is 3t, so it would take her 12 hours to do the same amount of work.
b
Let t be the time it takes Dominika to complete the work. Then, the time it would Heichi to complete the same amount of work would be t-6. The work in this case is painting 1 canoe, or 1. Now an expression to represent each person’s rate can be written using the formula r= Wt.
c|c Dominika's Rate & Heichi's Rate 1/t & 1/t-6 Heichi thinks that it would take them 4 hours to paint the canoe if they worked together. If their combined hourly rate, the sum of the individual ratios, is multiplied by 4, it would be 1, the number of canoes they could paint in 4 hours. 4 (1/t + 1/t-6 ) =1 Now solve the equation for t. First, eliminate the rational expressions in the denominators.
4 (1/t + 1/t-6 ) =1
▼
Rewrite
Distr
Distribute 4
4/t + 4/t-6 =1
MultEqn
LHS * t(t-6)=RHS* t(t-6)
t (t-6) ( 4/t + 4/t-6 ) = t(t-6)
Distr
Distribute t(t-6)
4t(t-6)/t + 4t(t-6)/t-6 = t(t-6)
CancelCommonFac
Cancel out common factors
4t(t-6)/t + 4t(t-6)/t-6 = t(t-6)
SimpQuot
Simplify quotient
4 (t-6) + 4t = t(t-6)
Distr
Distribute 4
4t-24 + 4t = t(t-6)
AddTerms
Add terms
8t-24 = t(t-6)
Distr
Distribute t
8t-24 = t^2 -6t
AddEqn
LHS+24=RHS+24
8t = t^2 -6t+24
SubEqn
LHS-8t=RHS-8t
0 = t^2 -14t +24
RearrangeEqn
Rearrange equation
t^2-14t+24 = 0
This quadratic equation can be solved by using the Quadratic Formula.
t^2-14t+24 = 0
▼
Solve using the quadratic formula
UseQuadForm
Use the Quadratic Formula: a = 1, b= - 14, c= 24
t=- ( - 14)±sqrt(( - 14)^2 -4* 1* 24)/2* 1
NegNeg
- (- a)=a
t=14±sqrt((- 14)^2 -4*1* 24)/2* 1
CalcPowProd
Calculate power and product
t=14±sqrt(196-96)/2
SubTerm
Subtract term
t=14±sqrt(100)/2
CalcRoot
Calculate root
t = 14 ± 10/2
The solutions for this equation are t = 14 ± 102, which can be separated and evaluated as follows.
| t = 14 ± 10/2 | |
|---|---|
| t = 14 + 10/2 | t = 14 - 10/2 |
| t = 12 | t = 2 |
Recall that t-6 is the time Heichi thinks it would take him to paint the canoe by himself. This value is negative when t=2, which does not make sense as time cannot be negative. This means that t must equal 12. Now confirm the solution in the original equation to check whether it is extraneous.
4 (1/t + 1/t-6 ) =1
Substitute
t= 12
4 (1/12 + 1/12-6 ) ? = 1
▼
Evaluate left-hand side
SubTerm
Subtract term
4 (1/12 + 1/6 ) ? = 1
ExpandFrac
a/b=a * 2/b * 2
4 (1/12 + 2/12 ) ? = 1
AddFrac
Add fractions
4 (3/12 ) ? = 1
Multiply
Multiply
12/12 ? = 1
CalcQuot
Calculate quotient
1 = 1 ✓
Substituting t=12 into the original equation resulted in an identity, so it is a valid solution to the equation. Therefore, according to Heichi, it would take Dominika 12 hours to paint the canoe by herself.
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Discussion
Rational Inequalities and Finding Their Solution Sets
A rational inequality is an inequality that contains a rational expression. x+4/x+3 + 1/x-5 > x-2/x-5 The general form of a rational inequality has a rational expression on the left-hand side of the inequality and 0 on the right-hand side of the inequality. x^2-2x-8/x-6 ≥ 0
The steps followed to solve a rational inequality are quite similar to those followed to solve linear inequalities.Method
Solving a Rational Inequality
To solve a rational inequality, the excluded values are first identified. The related rational equation is solved, then the excluded values and solutions of the equation are used to divide the number line into intervals. Finally, the solution set is determined by testing a value in each interval. Consider the following rational inequality. 4/x+1+x+1/x^2-4x-5 ≥ - 3/x-5 To solve this inequality, these steps can be followed.
1
Identify the Excluded Values
expand_more Take a look at the denominators of the rational expressions in the inequality.
4/x+1+x+1/x^2-4x-5 ≥ - 3/x-5
The excluded values are the ones that make the denominator 0.
| Denominator | Factored Form | Excluded Values |
|---|---|---|
| x+1 | - | x=- 1 |
| x^2-4x-5 | (x+1)(x-5) | x=- 1 and x=5 |
| x-5 | - | x= 5 |
The excluded values for this inequality are - 1 and 5.
2
Solve the Related Rational Equation
expand_more The related rational equation can be written by replacing the inequality sign with an equals sign.
Rational Inequality [0.6em]
4/x+1+x+1/x^2-4x-5 ≥ - 3/x-5 [1.2em] ⇓ [0.8em] Related Rational Equation [0.6em]
4/x+1+x+1/x^2-4x-5 = - 3/x-5
To solve this rational equation, all of the denominators must be eliminated. Notice that (x+1)(x-5) is the least common denominator.
4/x+1+x+1/x^2-4x-5 = - 3/x-5
▼
Simplify
MultEqn
LHS * (x+1)(x-5)=RHS* (x+1)(x-5)
(4/x+1+x+1/x^2-4x-5 )(x+1)(x-5) = (- 3/x-5 ) (x+1)(x-5)
Distr
Distribute (x+1)(x-5)
4(x+1)(x-5)/x+1+(x+1)(x+1)(x-5)/x^2-4x-5 = - 3(x+1)(x-5)/x-5
Substitute
x^2-4x-5= (x+1)(x-5)
4(x+1)(x-5)/x+1+(x+1)(x+1)(x-5)/(x+1)(x-5) = - 3(x+1)(x-5)/x-5
CancelCommonFac
Cancel out common factors
4(x+1)(x-5)/x+1+(x+1)(x+1)(x-5)/(x+1)(x-5) = - 3(x+1)(x-5)/x-5
SimpTerms
Simplify terms
4(x-5)+x+1= - 3(x+1)
Now that an equation without any denominators is obtained, it can be solved for x.
4(x-5)+x+1= - 3(x+1)
x=2
This is another value of interest.
3
Use the Values Found to Divide the Number Line Into Intervals
expand_more Next, the number line will be divided into intervals. The intervals are determined by the excluded values x=- 1 and x=5, and the solution to the related equation x=2.
4
Test a Value in Each Interval
expand_more Finally, a value in each interval will be tested to determine which intervals contain values that satisfy the inequality. Start with the interval x<- 1. The arbitrary point x=- 2 can be used. If substituting this point into the original inequality produces a true statement, then the numbers in this interval are solutions. If not, they are not solutions.
4/x+1+x+1/x^2-4x-5 ≥ - 3/x-5
Substitute
x= - 2
4/- 2+1+- 2+1/( - 2)^2-4( - 2)-5 ? ≥ - 3/- 2-5
▼
Evaluate left-hand side
CalcPow
Calculate power
4/-2+1+- 2+1/4-4(- 2)-5 ? ≥ - 3/- 2-5
MultNegNegOnePar
- a(- b)=a* b
4/-2 +1+- 2+1/4+8-5 ? ≥ - 3/- 2-5
AddSubTerms
Add and subtract terms
4/-1+- 1/7 ? ≥ - 3/- 2-5
ExpandFrac
a/b=a * 7/b * 7
28/-7+- 1/7 ? ≥ - 3/- 2-5
MoveNegDenomToNum
Put minus sign in numerator
- 28/7+- 1/7 ? ≥ - 3/- 2-5
AddFrac
Add fractions
- 29/7 ? ≥ - 3/- 2-5
- 29/7 ≱ 3/7 *
Solving for x=-2 resulted in a false statement. Therefore, the numbers less than - 1 are not solutions to the inequality. The same process can be applied to the other intervals to determine whether or not they are solutions to the inequality.
| x < - 1 | - 1 < x < 2 | 2 < x <5 | x > 5 | |
|---|---|---|---|---|
| Test Value | - 2 | 0 | 3 | 6 |
| Statement | - 29/7 ≱ 3/7 | 19/5 ≥ 3/5 | 1/2 ≱ 3/2 | 11/7 ≥ 3 |
The solutions to the given inequality are all the real numbers between - 1 and 2 or all the numbers greater than 5. (- 1, 2) or (5,∞) Notice that the given inequality is a non-strict inequality. Consequently, some of the endpoints might satisfy the inequality. The solution of the related equation, x=2, it also results in a true statement. Therefore, the square bracket is used for that endpoint to indicate that it is also a solution. Solution Set (- 1, 2] ⋃ (5,∞) This solution set can also be shown on the number line.
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Example
The Average Cost of a Basketball
The cost of producing x basketballs is represented by C(x). C(x) = 5x+2000
a
Find the average cost function, A(x).
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b
How many basketballs should be produced so that the average cost is less than $30? Write the answer as an inequality.
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Hint
a
Divide the cost function by x.
b
Start by writing a rational inequality. Then, solve the rational inequality. Do negative x-values make sense in this context?
Solution
a
The average cost function is found by dividing the cost function by the total number of items produced. The average cost function of producing x basketballs is then A(x) = C(x)x.
A(x) =C(x)/x ⇕ [0.8em] A(x) = 5x+2000/x
b
The values of x that make the average cost less than $30 need to be found. This can be represented algebraically by an inequality.
A(x) <30 ⇔ 5x+2000/x <30 To solve this rational inequality, these four steps will be followed.
- State the excluded values.
- Solve the related rational equation.
- Use the values determined in the previous steps to divide a number line into intervals.
- Test a value in each interval to determine which intervals contain values that satisfy the inequality.
Excluded Values
Consider the inequality. 5x+2000/x <30 The denominator is 0 only when x=0. The excluded value for this inequality is 0.
Solve the Related Equation
The related equation is obtained by replacing the inequality sign with an equals sign. c|c Rational Inequality & Related Equation [0.8em] 5x+2000/x < 30 & 5x+2000/x = 30 To solve the equation, multiply both sides by x.
5x+2000/x = 30
▼
Solve for x
MultEqn
LHS * x=RHS* x
5x+2000 = 30x
SubEqn
LHS-5x=RHS-5x
2000 = 25x
DivEqn
.LHS /25.=.RHS /25.
80 = x
RearrangeEqn
Rearrange equation
x=80
This value is another critical value.
Divide the Number Line Into Intervals
Next, the number line will be divided into intervals using the excluded value x=0 and the solution to the related equation x=80.
Test Values
Select an arbitrary value from each interval to see if it produces a true statement. Starting with the interval x<0, the point x=- 1 will be tested. If substituting this point into the inequality produces a true statement, then the numbers in this interval are solutions. If not, they are not solutions.
5x+2000/x < 30
Substitute
x= - 1
5( - 1)+2000/- 1 ? < 30
▼
Evaluate left-hand side
MultPosNeg
a(- b)=- a * b
- 5+2000/- 1 ? < 30
AddTerms
Add terms
1995/- 1 ? < 30
MoveNegDenomToFrac
Put minus sign in front of fraction
- 1995/1 ? < 30
DivByOne
a/1=a
- 1995 < 30 ✓
Since substituting x=-1 into the inequality resulted in a true statement, it can be concluded that the numbers less than 0 are solutions to the inequality. Use the same process for the other intervals to find any other solutions.
| x < 0 | 0 < x <80 | x >80 | |
|---|---|---|---|
| Choose a Test Value | - 1 | 1 | 100 |
| Statement | - 1995 <30 | 2005 < 30 | 25 < 30 |
The solutions to the inequality are all the numbers below 0 or above 80. Solution Set of Inequality x<0 or x> 80 However, x cannot be negative because it represents the number of basketballs produced. Therefore, that part of the solution can be ignored. Additionally, there is no need to check for endpoints as the inequality is strict. As a result, the average cost will be less than $30 as long as more than 80 basketballs are produced. x > 80
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Closure
Calculating Dominika’s Winning Percentages for Different Situations
When solving rational equations, the goal is to eliminate the rational denominators. However, the mechanics of solving rational inequalities are quite different. With all the methods discussed in this lesson, the challenge presented at the beginning can be reconsidered now. Dominika and Heichi keep track of how many games each of them win.
Given that Dominika has won 24 out of the 40 games against Heichi, the following situation will be analyzed.
a
How many games would Dominika have to win in a row to have a 75 % winning record?
b
How many games would Dominika have to win in a row to have a 90 % winning record?
c
Is Dominika able to reach a 100 % winning record? Explain why or why not.
d
Suppose that after reaching a winning record of 90 %, Dominika had a losing streak. How many games in a row would Dominika have to lose to drop down to a winning record below 50 % again?
Answer
a
24 games
b
120 games
c
No, see solution.
d
More than 128 games
Hint
a
Let x be the number of games that Dominika wins. Write a proportion using x.
b
Repeat the same procedure as in the previous section.
c
Is there a scenario where Dominika wins all the games?
d
Consider the answer found in Part B. Write a rational inequality for the number of games that Dominika loses without winning any.
Solution
a
The ratio of Dominika's wins to the total number of games is currently 2440.
24/40 = 0.6 ⇔ 60 % Dominika wants to win enough games to bring this percentage up to 75 %. Let x be the least number of games that Dominika has to win. Then, x must be added to both the numerator and denominator, making the new ratio 0.75 24 +x/40 +x = 0.75 Now solve this rational equation by multiplying both sides by 40+x and simplifying.
If Dominika wins the next 24 games, her wins to losses percentage will be 75 %. Note that this solution is not extraneous.
b
The number of games x that Dominika has to win can be found by following the same process as in the previous part. In this case, Dominika wants to win 90 % of the games. Therefore, the ratio of 24+x to 40+x should be equal to 0.90.
Dominika has to win the next 120 games in order to bring her percentage of wins up to 90 %. This solution is also not an extraneous solution.
c
In order for Dominika to have won 100 % of the games, she must have won all of them. However, this is not the case because she has already lost 16 out of the 40 games played.
| Dominika | |
|---|---|
| Win | Loss |
| 24 | 16 |
This fact cannot be changed, no matter how many more games she wins. Try following the same procedure as the previous sections to see what happens. 24 + x/40+x = 1 ⇓ 24 + x = 40 + x ⇓ 24 = 40 * Attempting to solve this equation results in a false statement. This means that there is no value of x that will bring Dominika's record up to 100 %.
d
Earlier it was determined that Dominika has to win 120 consecutive games in order to bring her winning record up to 90 %. In other words, Dominika will have to win 144 out of 160 games.
20+120/40+120 = 144/160 From this point, if she were to lose x consecutive games, her record would be 144 wins out of 160+x games. To find the value to bring her record down to below 0.50, the following inequality must be solved. 144/160+x < 0.50 The excluded value for this inequality is - 160, as this value makes the denominator equal to zero. Next, the related rational equation will be solved.
144/160+x = 0.50
▼
Solve for x
MultEqn
LHS * 160+x=RHS* 160+x
144 = 0.50(160+x)
Distr
Distribute 0.50
144 = 80 + 0.50x
SubEqn
LHS-80=RHS-80
64 = 0.50x
DivEqn
.LHS /0.5.=.RHS /0.5.
128 = x
RearrangeEqn
Rearrange equation
x=128
The excluded value x=- 160 and the solution to the related equation x=128 will be used to divide the number line into intervals.
Finally, a value from each interval is chosen and tested. If substituting this point into the inequality produces a true statement, then the numbers in this interval are solutions. If not, these numbers are not solutions.
Therefore, the solutions to the inequality are all the numbers below - 160 and above 128. Solution Set of Inequality x< - 160 or x > 128 Notice that x cannot be negative because it is a number of games. Dominika cannot lose a negative number of games. As such, that part of the solution can be ignored. Additionally, there is no need to check for endpoints because the inequality is strict. If Dominika loses more than the next 128 games, she will have won less than 50 % of the total games she and Heichi played. x > 128
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Solving Rational Equations and Inequalities
Exercise 2.1
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To find the work done by Ramsha when working together with her sister Maya, we can use the formula below. ( Work done)=( Work rate)*( Time) We know that Ramsha can clean a room in 5 hours when working alone. Therefore, her work rate is 15 room/hours. We also know that Ramsha and Maya can clean the room in 3 hours when working together. We can find the work done by Ramsha when working together with Maya by substituting these values into the formula.
The work done by Ramsha is 35 room when working together with Maya.
Ramsha and Maya do 100 % of the work when they work together. Therefore, the sum of the work done by Ramsha and Maya is 100 % = 1. We found in Part A that the work done by Ramsha is 35 room when working together with Maya. Therefore, we can find the work done d by Maya by solving the following equation.
3/5+d= 1
Now, we will solve the equation for d.
The work done by Maya when working together with Ramsha is 25. Next, using the formula in Part A, we can find the work rate of Maya.
The work rate can be expressed as 1t room/hours. Work rate=1/t ⇒ 1/t=2/15 Finally, by solving this equation for t, we can find how long Maya would take to clean the room when working alone.
It takes 7.5 hours for Maya to clean the room when working alone.
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We want to solve the given rational equation. 3/x-4 + 1/x = x-1/x-4 To do so, we will first determine the least common denominator (LCD) so that we can clear the denominators. Note that the denominators are already factored. Therefore, in this case the LCD is the product of (x-4) and x.
We have a quadratic equation in standard form. Let's identify the values of a, b, and c. x^2 -5x+4=0 ⇔ 1x^2 +( -5x)+ 4=0 We can see that a= 1, b= -5, and c= 4. Let's substitute these values into the Quadratic Formula to find the possible values of x.
Let's calculate both solutions by using the positive and negative signs.
| x=5 ± 3/2 | |
|---|---|
| x_1=5+ 3/2 | x_2=5- 3/2 |
| x_1=8/2 | x_2=2/2 |
| x_1=4 | x_2=1 |
Finally, we will check whether either x=4 or x=1 are extraneous solutions. To do so, we will substitute 4 and 1 for x in the original equation. Let's start with the first one.
Since the denominator cannot be 0, x=3 is an extraneous solution. Now, we will check our second solution, x=1.
The only solution to the given equation is x=1.
We need to graph the given rational functions to determine which statements are correct.
y_1 = 3/x-4+1/x and y_2 = x-1/x-4
Let's take a look at their graphs drawn on the same coordinate plane.
We see that the graphs of the functions intersect at x=1. For the extraneous solution of x=4, the graphs do not intersect. In fact, x=4 is a vertical asymptote for both functions. There is no asymptote at x=3.
We can conclude that I, II, and IV are the correct ones among the given statements. As a general result, if the graphs intersect, then we can say that the point of intersection is a solution. If the graphs do not intersect, then the possible solution is extraneous.
We will begin by graphing y= 3x-4+ 1x. Let's first simplify it!
Now, we can find the asymptotes of the function. To find the vertical asymptotes, we will solve x(x-4)=0 by using the Zero Product Property.
Therefore, the function has vertical asymptotes at x=0 and x=4. Next, we will find the horizontal asymptote of the function. To do so, let's pay close attention to the degrees of the numerator and denominator. y=4x^1-4/x^2-4x We see that the degree of the denominator is higher than the degree of the numerator. Therefore, the line y=0 is a horizontal asymptote. Now, we will plot the asymptotes.
Next, we will make a table of values. Recall that we are asked to graph the function for x-values greater than 0 and less than 7. Since the function has a vertical asymptote at x=4, this value will be excluded from the domain.
| x | 4x-4/x(x-4) | y |
|---|---|---|
| 0.5 | 4( 0.5)-4/0.5( 0.5-4) | ≈ 1.1 |
| 1 | 4( 1)-4/1( 1-4) | 0 |
| 2 | 4( 2)-4/2( 2-4) | - 1 |
| 3 | 4( 3)-4/3( 3-4) | ≈ -2.7 |
| 3.5 | 4( 3.5)-4/3.5( 3.5-4) | ≈ -5.7 |
| 4.5 | 4( 4.5)-4/4.5( 4.5-4) | ≈ 6.2 |
| 5 | 4( 5)-4/5( 5-4) | 3.2 |
| 6 | 4( 6)-4/6( 6-4) | ≈ 1.7 |
Finally, let's plot and connect the points.
We can graph y= x-1x-4 proceeding in the same way. The function has also a vertical asymptote at x=4. Let's pay close attention to the degrees of the numerator and denominator. y=x^1-1/x^1-4 We see that the degrees of the numerator and denominator are the same. To find the horizontal asymptote, we need to find the quotient between the leading coefficients. y=1x-1/1x-3 Since 11=1, there is a horizontal asymptote at y=1. By finding, plotting, and connecting points, we can graph the function as we did for the first one.
| x | x-1/x-4 | y |
|---|---|---|
| 0.5 | 0.5-1/0.5-4 | ≈ 0.1 |
| 1 | 1-1/1-4 | 0 |
| 2 | 2-1/2-4 | - 0.5 |
| 3 | 3-1/3-4 | - 2 |
| 3.5 | 3.5-1/3.5-4 | - 5 |
| 4.5 | 4.5-1/4.5-4 | 7 |
| 5 | 5-1/5-4 | 4 |
| 6 | 6-1/6-4 | 2.5 |
Let's draw it!
We can now plot the graphs on the same coordinate plane.
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Consider the set of rational functions with the following form. 1/x = ax+b, wherea ≠ 0 Which of the following can be the number of solutions to an equation in this form?
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We want to determine the number of solutions of an equation of the following form. 1/x = ax+b, a ≠ 0 We can write two functions from the equation, one function for each side. y = 1/x and y = ax+b For simplicity, let's consider the graph of y= 1x and y=x.
We see that there are two points of intersection for these functions. We can also find lines that intersect the given function at one point or that do not intersect at any point.
Each of these intersection points is a solution of the equation. Let's now use an interactive graph of the two functions to check if there is another possible case. Change the values of a and b.
By observing the applet, we can see that there are indeed three cases. We can generalize these cases as follows.
- There are no solutions when the slope a of the line is negative and b=0.
- There is exactly one solution when the line is tangent to the graph of y= 1x.
- There are exactly two solutions when the slope a of the line is positive.
Therefore, we can conclude that a rational equation in the form 1x=ax+b can have exactly one solution, exactly two solutions, or no solution.
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Solving Rational Equations and Inequalities
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