Mastering Rational Equations and Inequalities: Comprehensive Guide

Distribute - 1

- x^2+2 = (x-3)(- x+4)

▼

Rewrite

Distribute (x-3)

- x^2+2 = (x-3)(- x) + (x-3)(4)

Distribute - x

- x^2+2 = - x^2+3x + (x-3)(4)

Distribute 4

- x^2+2 = - x^2+3x + 4x-12

Add terms

- x^2+2 = - x^2+7x-12

▼

Solve for x

LHS+x^2=RHS+x^2

2 = 7x-12

LHS+12=RHS+12

14=7x

.LHS /7.=.RHS /7.

2 =x

Rearrange equation

x = 2

The value that satisfies the equation is 2. Next, check if it also satisfies the original equation.

Check for Extraneous Solutions

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

x= 2

- 1/2-3 ? = - 2+4/2^2-2

▼

Evaluate left-hand side

Subtract term

- 1/- 1 ? = - 2+4/2^2-2

Calculate quotient

1 ? = - 2+4/2^2-2

▼

Evaluate right-hand side

CalcPow

Calculate power

1 ? = - 2+4/4-2

Add and subtract terms

1 ? = 2/2

Calculate quotient

1 = 1 ✓

Substiting x=2 in the original equation produced a true statement. Therefore, it is a solution to the original equation.

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?

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

45/100=2.25+0.6x/15+x

CrossMult

Cross multiply

45(15+x)=100(2.25+0.6x)

▼

Solve for x

Distribute 45

675+45x=100(2.25+0.6x)

Distribute 100

675+45x=225+60x

LHS-60x=RHS-60x

625-15x=225

LHS-675=RHS-675

- 15x= - 450

.LHS /(- 15).=.RHS /(- 15).

x = 30

The solution might be extraneous, so it must be verified in the original equation.

45/100=0.15(15)+0.6(x)/15+x

x= 30

45/100 ? = 0.15(15)+0.6( 30)/15+ 30

▼

Evaluate right-hand side

45/100 ? = 2.25+18/15+30

Add terms

45/100 ? = 20.25/45

Calculate quotient

0.45 = 0.45 ✓

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.

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.

Find the Least Common Denominator

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

▼

Factor

WriteSum

Write as a sum

x^2+2x+5x+10

Factor out x

x(x+2)+5x+10

Factor out 5

x(x+2)+ 5(x+2)

Factor out (x+2)

(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)

Multiply the Equation by the Least Common Denominator

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

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

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

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

Simplify quotient

5(x+5)-4(x+2) = 16

Solve the Resulting Equation

Now the resulting equation can be solved by using the Distributive Property and inverse operations.

5(x+5)-4(x+2) = 16

Distribute 5

5x+25-4(x+2) = 16

Distribute - 4

5x+25-4x-8 = 16

SubTerms

Subtract terms

x + 17 = 1 6

LHS-17=RHS-17

x = - 1

Check for Extraneous Solutions

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

x= - 1

5/- 1+2 - 4/- 1+5 ? = 16/( - 1)^2+7( - 1)+10

▼

Evaluate left-hand side

Add terms

5/1 - 4/4? = 16/(- 1)^2+7(- 1)+10

DivByOne

a/1=a

5 -4/4? =16/(- 1)^2+7(- 1)+10

Calculate quotient

5-1 ? = 16/(- 1)^2+7(- 1)+10

Subtract term

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

Add and subtract terms

4 ? = 16/4

Calculate quotient

4 = 4 ✓

The value made a true statement. Therefore, it is a solution to the equation.

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.

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

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

Distribute (x+3)(x-3)

(x+3)(x-3)/x+3 + (x+3)(x-3)/x-3 = (x+3)(x-3)1.5

Cancel out common factors

(x+3)(x-3)/x+3 + (x+3)(x-3)/x-3 = (x+3)(x-3)1.5

Simplify quotient

x-3+x+3 = (x+3)(x-3)1.5

Add terms

2x = (x+3)(x-3)1.5

▼

Simplify right-hand side

ExpandDiffSquares

(a+b)(a-b)=a^2-b^2

2x = (x^2-9)1.5

Distribute 1.5

2x = 1.5x^2-13.5

LHS-2x=RHS-2x

0 = 1.5x^2-2x-13.5

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

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

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

Factor out 2

x_(1,2) = 2(2 ± sqrt(85))/6

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

x= 3.74

1/3.74+3 + 1/3.74-3 ? = 1.5

▼

Evaluate left-hand side

Add terms

1/6.74 + 1/3.74-3 ? = 1.5

Subtract term

1/6.74 + 1/0.74 ? = 1.5

a/b=a * 0.74/b * 0.74

0.74/4.9876 + 1/0.74 ? = 1.5

a/b=a * 6.74/b * 6.74

0.74/4.9876 + 6.74/4.9876? = 1.5

Add fractions

7.48/4.9876 ? = 1.5

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.

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

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

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

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

Simplify quotient

2-(x+1)=x-1

Distribute - 1

2-x-1 = x-1

LHS+1=RHS+1

2-x = x

LHS+x=RHS+x

2 = 2x

.LHS /2.=.RHS /2.

1 = x

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.

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?

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?

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

a/b=a * 3/b * 3

3 (1/3t + 3/3t ) =1

Add fractions

3 (4/3t ) =1

12/3t = 1

LHS * 3t=RHS* 3t

12 = 3t

.LHS /3.=.RHS /3.

4 = t

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

t= 4

3 (1/3( 4) + 1/4 ) ? = 1

▼

Evaluate left-hand side

3 (1/12 + 1/4 ) ? = 1

a/b=a * 3/b * 3

3 (1/12 + 3/12 ) ? = 1

Add fractions

3 (4/12 ) ? = 1

12/12 ? = 1

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

Distribute 4

4/t + 4/t-6 =1

LHS * t(t-6)=RHS* t(t-6)

t (t-6) ( 4/t + 4/t-6 ) = t(t-6)

Distribute t(t-6)

4t(t-6)/t + 4t(t-6)/t-6 = t(t-6)

Cancel out common factors

4t(t-6)/t + 4t(t-6)/t-6 = t(t-6)

Simplify quotient

4 (t-6) + 4t = t(t-6)

Distribute 4

4t-24 + 4t = t(t-6)

Add terms

8t-24 = t(t-6)

Distribute t

8t-24 = t^2 -6t

LHS+24=RHS+24

8t = t^2 -6t+24

LHS-8t=RHS-8t

0 = t^2 -14t +24

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

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

t= 12

4 (1/12 + 1/12-6 ) ? = 1

▼

Evaluate left-hand side

Subtract term

4 (1/12 + 1/6 ) ? = 1

a/b=a * 2/b * 2

4 (1/12 + 2/12 ) ? = 1

Add fractions

4 (3/12 ) ? = 1

12/12 ? = 1

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.

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.

Identify the Excluded Values

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.

Solve the Related Rational Equation

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

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)

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

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

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)

▼

Solve for x

Distribute 4

4x-20+x+1= - 3(x+1)

Distribute - 3

4x-20+x+1= - 3x-3

Add and subtract terms

5x-19 = - 3x-3

LHS+3x=RHS+3x

8x-19 = - 3

LHS+19=RHS+19

8x = 16

.LHS /8.=.RHS /8.

x=2

This is another value of interest.

Use the Values Found to Divide the Number Line Into Intervals

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.

Test a Value in Each Interval

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

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

Add and subtract terms

4/-1+- 1/7 ? ≥ - 3/- 2-5

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

Add fractions

- 29/7 ? ≥ - 3/- 2-5

▼

Evaluate right-hand side

Subtract term

- 29/7 ? ≥ - 3/- 7

DivNegNeg

- a/- b=a/b

- 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.

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).

b How many basketballs should be produced so that the average cost is less than $30? Write the answer as an inequality.

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

LHS * x=RHS* x

5x+2000 = 30x

LHS-5x=RHS-5x

2000 = 25x

.LHS /25.=.RHS /25.

80 = x

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

x= - 1

5( - 1)+2000/- 1 ? < 30

▼

Evaluate left-hand side

MultPosNeg

a(- b)=- a * b

- 5+2000/- 1 ? < 30

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

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.

24+x/40+x = 0.75

LHS * (40+x)=RHS* (40+x)

24+x = 0.75(40+x)

▼

Solve for x

Distribute 0.75

24+x = 30+0.75x

LHS-24=RHS-24

x = 6 + 0.75x

LHS-0.75x=RHS-0.75x

0.25x = 6

.LHS /0.25.=.RHS /0.25.

x = 24

If Dominika wins the next 24 games, her wins to losses percentage will be 75 %. Note that this solution is not extraneous.

24+x/40+x = 0.75

x= 24

24+ 24/40+ 24 ? = 0.75

▼

Evaluate left-hand side

Add terms

48/64 ? = 0.75

Calculate quotient

0.75 = 0.75 ✓

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.

24+x/40+x = 0.90

LHS * (40+x)=RHS* (40+x)

24+x = 0.90(40+x)

▼

Solve for x

Distribute 0.90

24+x = 36+0.90x

LHS-24=RHS-24

x = 12 + 0.90x

LHS-0.90x=RHS-0.90x

0.10x = 12

.LHS /0.10.=.RHS /0.10.

x = 120

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.

24+x/40+x = 0.90

x= 120

24+ 120/40+ 120 ? = 0.90

▼

Evaluate left-hand side

Add terms

144/160 ? = 0.90

Calculate quotient

0.90 = 0.90 ✓

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

LHS * 160+x=RHS* 160+x

144 = 0.50(160+x)

Distribute 0.50

144 = 80 + 0.50x

LHS-80=RHS-80

64 = 0.50x

.LHS /0.5.=.RHS /0.5.

128 = x