Mastering Rational Equations and Inequalities: Comprehensive Guide

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.

Cross Products Property
Least Common Denominator
Inequality
Solution Set of an Inequality

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?

Discussion

Rational Equations and Finding Their Solutions

A rational equation is an equation that contains at least one rational expression.

\frac{1}{x} + \frac{x}{2} = \frac{x + 4}{2 x}

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
$\frac{x}{2 x + 7} = \frac{2}{x - 1}$	Cross Products Property
$\frac{x}{x - 5} + \frac{x}{x + 5} = \frac{2}{x ^{2} - 2 5}$	LCD

When a rational equation has been solved using algebraic methods, it is necessary to ensure that each solution satisfies the original equation because extraneous solutions may appear.

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.

\frac{- 1}{x - 3} = \frac{- x + 4}{x ^{2} - 2}

The equation is basically a proportion. Therefore, it can be solved by using the Cross Products Property.

Use the Cross Products Property

To get the variable terms out of the denominators, the Cross Product Property will be used.

\frac{- 1}{x - 3} = \frac{- x + 4}{x ^{2} - 2} ⇓ - 1 \cdot (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 .

Solve the Resulting Equation

To solve the resulting equation, use the Distributive Property and combine like terms.

- 1 \cdot (x^{2} - 2) = (x - 3) (- x + 4)

Distr

Distribute $- 1$

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

▼

Rewrite

Distr

Distribute $(x - 3)$

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

Distr

Distribute $- x$

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

Distr

Distribute $4$

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

AddTerms

Add terms

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

▼

Solve for

x

AddEqn

$LHS + x^{2} = RHS + x^{2}$

2 = 7 x - 12

AddEqn

$LHS + 12 = RHS + 12$

14 = 7 x

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.

Check for Extraneous Solutions

The solution found in the previous step might be extraneous. It must be verified in the original equation.

\frac{- 1}{x - 3} = \frac{- x + 4}{x ^{2} - 2}

Substitute

$x = 2$

\frac{- 1}{2 - 3} = ? \frac{- 2 + 4}{2 ^{2} - 2}

▼

Evaluate left-hand side

SubTerm

Subtract term

\frac{- 1}{- 1} = ? \frac{- 2 + 4}{2 ^{2} - 2}

CalcQuot

Calculate quotient

1 = ? \frac{- 2 + 4}{2 ^{2} - 2}

▼

Evaluate right-hand side

CalcPow

Calculate power

1 = ? \frac{- 2 + 4}{4 - 2}

AddSubTerms

Add and subtract terms

1 = ? \frac{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.

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 = \frac{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.

\frac{4 5}{1 0 0} = \frac{0 . 1 5 ( 1 5 ) + 0 . 6 ( x )}{1 5 + x}

This proportion is also a rational expression. To solve the equation for

x,

the Cross Products Property can be used.

\frac{4 5}{1 0 0} = \frac{0 . 1 5 ( 1 5 ) + 0 . 6 ( x )}{1 5 + x}

Multiply

\frac{4 5}{1 0 0} = \frac{2 . 2 5 + 0 . 6 x}{1 5 + x}

CrossMult

Cross multiply

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

▼

Solve for

x

Distr

Distribute $45$

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

Distr

Distribute $100$

675 + 45 x = 225 + 60 x

SubEqn

$LHS - 60 x = RHS - 60 x$

625 - 15 x = 225

SubEqn

$LHS - 675 = RHS - 675$

- 15 x = - 450

DivEqn

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

x = 30

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

\frac{4 5}{1 0 0} = \frac{0 . 1 5 ( 1 5 ) + 0 . 6 ( x )}{1 5 + x}

Substitute

$x = 30$

\frac{4 5}{1 0 0} = ? \frac{0 . 1 5 ( 1 5 ) + 0 . 6 ( 3 0 )}{1 5 + 3 0}

▼

Evaluate right-hand side

Multiply

\frac{4 5}{1 0 0} = ? \frac{2 . 2 5 + 1 8}{1 5 + 3 0}

AddTerms

Add terms

\frac{4 5}{1 0 0} = ? \frac{2 0 . 2 5}{4 5}

CalcQuot

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.

\frac{5}{x + 2} - \frac{4}{x + 5} = \frac{1 6}{x ^{2} + 7 x + 1 0}

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.

\frac{5}{x + 2} - \frac{4}{x + 5} = \frac{1 6}{x ^{2} + 7 x + 1 0}

The denominator of the rational expression on the left-hand side cannot be factored further. Factor the denominator of the other expression.

x^{2} + 7 x + 10

▼

Factor

WriteSum

Write as a sum

x^{2} + 2 x + 5 x + 10

FactorOut

Factor out $x$

x (x + 2) + 5 x + 10

FactorOut

Factor out $5$

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

FactorOut

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
$\frac{5}{x + 2}$	$(x + 2) (x + 5)$
$\frac{4}{x + 5}$
$\frac{1 6}{( x + 2 ) ( x + 5 )}$

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.

\frac{5}{x + 2} - \frac{4}{x + 5} = \frac{1 6}{x ^{2} + 7 x + 1 0}

MultEqn

$LHS \cdot (x + 2) (x + 5) = RHS \cdot (x + 2) (x + 5)$

(x + 2) (x + 5) (\frac{5}{x + 2} - \frac{4}{x + 5}) = (x + 2) (x + 5) (\frac{1 6}{x ^{2} + 7 x + 1 0})

▼

Simplify

Distr

Distribute $(x + 2) (x + 5)$

\frac{5 ( x + 2 ) ( x + 5 )}{x + 2} - \frac{4 ( x + 2 ) ( x + 5 )}{x + 5} = \frac{1 6 ( x + 2 ) ( x + 5 )}{x ^{2} + 7 x + 1 0}

CancelCommonFac

Cancel out common factors

\frac{5 ( x + 2 ) ( x + 5 )}{x + 2} - \frac{4 ( x + 2 ) ( x + 5 )}{x + 5} = \frac{1 6 ( x + 2 ) ( x + 5 )}{x ^{2} + 7 x + 1 0}

SimpQuot

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

Distr

Distribute $5$

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

Distr

Distribute $- 4$

5 x + 25 - 4 x - 8 = 16

SubTerms

Subtract terms

x + 17 = 16

SubEqn

$LHS - 17 = RHS - 17$

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.

\frac{5}{x + 2} - \frac{4}{x + 5} = \frac{1 6}{x ^{2} + 7 x + 1 0}

Substitute

$x = - 1$

\frac{5}{- 1 + 2} - \frac{4}{- 1 + 5} = ? \frac{1 6}{( - 1 ) ^{2} + 7 ( - 1 ) + 1 0}

▼

Evaluate left-hand side

AddTerms

Add terms

\frac{5}{1} - \frac{4}{4} = ? \frac{1 6}{( - 1 ) ^{2} + 7 ( - 1 ) + 1 0}

DivByOne

$\frac{a}{1} = a$

5 - \frac{4}{4} = ? \frac{1 6}{( - 1 ) ^{2} + 7 ( - 1 ) + 1 0}

CalcQuot

Calculate quotient

5 - 1 = ? \frac{1 6}{( - 1 ) ^{2} + 7 ( - 1 ) + 1 0}

SubTerm

Subtract term

4 = ? \frac{1 6}{( - 1 ) ^{2} + 7 ( - 1 ) + 1 0}

▼

Evaluate right-hand side

CalcPow

Calculate power

4 = ? \frac{1 6}{1 + 7 ( - 1 ) + 1 0}

MultPosNeg

$a (- b) = - a \cdot b$

4 = ? \frac{1 6}{1 - 7 + 1 0}

AddSubTerms

Add and subtract terms

4 = ? \frac{1 6}{4}

CalcQuot

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 = r t .

This formula can be maniuplated to represent distance in terms of time.

d = r t \Leftrightarrow t = \frac{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

\frac{1}{x + 3}

hours for the outbound portion and

\frac{1}{x - 3}

hours for the return portion.

	Going	Returning
Distance (km)	$1$	$1$
Speed (km/h)	$x + 3$	$x - 3$
Time (h)	$\frac{1}{x + 3}$	$\frac{1}{x - 3}$

The sum of the portions is equal to the total round trip time.

\frac{1}{x + 3} + \frac{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.

\frac{1}{x + 3} + \frac{1}{x - 3} = 1.5

MultEqn

$LHS \cdot (x + 3) (x - 3) = RHS \cdot (x + 3) (x - 3)$

(x + 3) (x - 3) (\frac{1}{x + 3} + \frac{1}{x - 3}) = (x + 3) (x - 3) 1.5

▼

Simplify left-hand side

Distr

Distribute $(x + 3) (x - 3)$

\frac{( x + 3 ) ( x - 3 )}{x + 3} + \frac{( x + 3 ) ( x - 3 )}{x - 3} = (x + 3) (x - 3) 1.5

CancelCommonFac

Cancel out common factors

\frac{( x + 3 ) ( x - 3 )}{x + 3} + \frac{( 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

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

▼

Simplify right-hand side

ExpandDiffSquares

$(a + b) (a - b) = a^{2} - b^{2}$

2 x = (x^{2} - 9) 1.5

Distr

Distribute $1.5$

2 x = 1.5 x^{2} - 13.5

SubEqn

$LHS - 2 x = RHS - 2 x$

0 = 1.5 x^{2} - 2 x - 13.5

RearrangeEqn

Rearrange equation

1.5 x^{2} - 2 x - 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.

1.5 x^{2} - 2 x - 13.5 = 0

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

3 x^{2} - 4 x - 27 = 0

▼

Solve using the quadratic formula

UseQuadForm

Use the Quadratic Formula: $a = 3, b = - 4, c = - 27$

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( 3 ) ( - 2 7 )}{2 ( 3 )}

NegNeg

$- (- a) = a$

x_{1, 2} = \frac{4 \pm ( - 4 ) ^{2} - 4 ( 3 ) ( - 2 7 )}{2 ( 3 )}

CalcPowProd

Calculate power and product

x_{1, 2} = \frac{4 \pm 1 6 + 3 2 4}{6}

AddTerms

Add terms

x_{1, 2} = \frac{4 \pm 3 4 0}{6}

SplitIntoFactors

Split into factors

x_{1, 2} = \frac{4 \pm 4 \cdot 8 5}{6}

SqrtProd

$a \cdot b = a \cdot b$

x_{1, 2} = \frac{4 \pm 4 \cdot 8 5}{6}

CalcRoot

Calculate root

x_{1, 2} = \frac{4 \pm 2 \cdot 8 5}{6}

FactorOut

Factor out $2$

x_{1, 2} = \frac{2 ( 2 \pm 8 5 )}{6}

SimpQuot

Simplify quotient

x = \frac{2 \pm 8 5}{3}

The solutions for this equation are

x = \frac{2 \pm 8 5}{3} .

Separate them into the positive and negative cases and round the answers to two decimal places.

$x = \frac{2 \pm 8 5}{3}$
$x_{1} = \frac{2 + 8 5}{3}$	$x_{2} = \frac{2 - 8 5}{3}$
$x_{1} = \frac{2}{3} + \frac{8 5}{3}$	$x_{2} = \frac{2}{3} - \frac{8 5}{3}$
$x_{1} \approx 3.74$	$x_{2} \approx - 2.41$

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 .

\frac{1}{x + 3} + \frac{1}{x - 3} = ? 1.5

Substitute

$x = 3.74$

\frac{1}{3 . 7 4 + 3} + \frac{1}{3 . 7 4 - 3} = ? 1.5

▼

Evaluate left-hand side

AddTerms

Add terms

\frac{1}{6 . 7 4} + \frac{1}{3 . 7 4 - 3} = ? 1.5

SubTerm

Subtract term

\frac{1}{6 . 7 4} + \frac{1}{0 . 7 4} = ? 1.5

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 0 . 7 4}{b \cdot 0 . 7 4}$

\frac{0 . 7 4}{4 . 9 8 7 6} + \frac{1}{0 . 7 4} = ? 1.5

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 6 . 7 4}{b \cdot 6 . 7 4}$

\frac{0 . 7 4}{4 . 9 8 7 6} + \frac{6 . 7 4}{4 . 9 8 7 6} = ? 1.5

AddFrac

Add fractions

\frac{7 . 4 8}{4 . 9 8 7 6} = ? 1.5

CalcQuot

Calculate quotient

1.499719 \dots = ? 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.

\frac{2}{x ^{2} - 1} - \frac{1}{x - 1} = \frac{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),

x^{2} - 1

is the common denominator for the three rational expressions. Multiply the equation by this common denominator and solve for

x .

\frac{2}{x ^{2} - 1} - \frac{1}{x - 1} = \frac{1}{x + 1}

MultEqn

$LHS \cdot (x - 1) (x + 1) = RHS \cdot (x - 1) (x + 1)$

(x - 1) (x + 1) (\frac{2}{x ^{2} - 1} - \frac{1}{x - 1}) = (x - 1) (x + 1) (\frac{1}{x + 1})

▼

Solve for

x

Distr

Distribute $(x - 1) (x + 1)$

\frac{2 ( x - 1 ) ( x + 1 )}{x ^{2} - 1} - \frac{( x - 1 ) ( x + 1 )}{x - 1} = (x - 1) (x + 1) (\frac{1}{x + 1})

MoveLeftFacToNumOne

$a \cdot \frac{1}{b} = \frac{a}{b}$

\frac{2 ( x - 1 ) ( x + 1 )}{x ^{2} - 1} - \frac{( x - 1 ) ( x + 1 )}{x - 1} = \frac{( x - 1 ) ( x + 1 )}{x + 1}

FacDiffSquares

$a^{2} - b^{2} = (a + b) (a - b)$

\frac{2 ( x - 1 ) ( x + 1 )}{( x - 1 ) ( x + 1 )} - \frac{( x - 1 ) ( x + 1 )}{x - 1} = \frac{( x - 1 ) ( x + 1 )}{x + 1}

CancelCommonFac

Cancel out common factors

\frac{2 ( x - 1 ) ( x + 1 )}{( x - 1 ) ( x + 1 )} - \frac{( x - 1 ) ( x + 1 )}{x - 1} = \frac{( 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 = 2 x

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 .

\frac{2}{0 1 ^{2} - 1} - \frac{1}{0 1 - 1 ∣} = \frac{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

x = - 1 .

When

x = 1

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.

\underline{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

3 t .

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 = \frac{W}{t} .

Dominika ’ s Rate \frac{1}{3 t} Heichi ’ s Rate \frac{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 (\frac{1}{3 t} + \frac{1}{t}) = 1

Now solve the rational equation for

x .

3 (\frac{1}{3 t} + \frac{1}{t}) = 1

▼

Solve for

t

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 3}{b \cdot 3}$

3 (\frac{1}{3 t} + \frac{3}{3 t}) = 1

AddFrac

Add fractions

3 (\frac{4}{3 t}) = 1

Multiply

\frac{1 2}{3 t} = 1

MultEqn

$LHS \cdot 3 t = RHS \cdot 3 t$

12 = 3 t

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 (\frac{1}{3 t} + \frac{1}{t}) = 1

Substitute

$t = 4$

3 (\frac{1}{3 ( 4 )} + \frac{1}{4}) = ? 1

▼

Evaluate left-hand side

Multiply

3 (\frac{1}{1 2} + \frac{1}{4}) = ? 1

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 3}{b \cdot 3}$

3 (\frac{1}{1 2} + \frac{3}{1 2}) = ? 1

AddFrac

Add fractions

3 (\frac{4}{1 2}) = ? 1

Multiply

\frac{1 2}{1 2} = ? 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

3 t,

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 = \frac{W}{t} .

Dominika ’ s Rate \frac{1}{t} Heichi ’ s Rate \frac{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 (\frac{1}{t} + \frac{1}{t - 6}) = 1

Now solve the equation for

t .

First, eliminate the rational expressions in the denominators.

4 (\frac{1}{t} + \frac{1}{t - 6}) = 1

▼

Rewrite

Distr

Distribute $4$

\frac{4}{t} + \frac{4}{t - 6} = 1

MultEqn

$LHS \cdot t (t - 6) = RHS \cdot t (t - 6)$

t (t - 6) (\frac{4}{t} + \frac{4}{t - 6}) = t (t - 6)

Distr

Distribute $t (t - 6)$

\frac{4 t ( t - 6 )}{t} + \frac{4 t ( t - 6 )}{t - 6} = t (t - 6)

CancelCommonFac

Cancel out common factors

\frac{4 t ( t - 6 )}{t} + \frac{4 t ( t - 6 )}{t - 6} = t (t - 6)

SimpQuot

Simplify quotient

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

Distr

Distribute $4$

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

AddTerms

Add terms

8 t - 24 = t (t - 6)

Distr

Distribute $t$

8 t - 24 = t^{2} - 6 t

AddEqn

$LHS + 24 = RHS + 24$

8 t = t^{2} - 6 t + 24

SubEqn

$LHS - 8 t = RHS - 8 t$

0 = t^{2} - 14 t + 24

RearrangeEqn

Rearrange equation

t^{2} - 14 t + 24 = 0

This quadratic equation can be solved by using the Quadratic Formula.

t^{2} - 14 t + 24 = 0

▼

Solve using the quadratic formula

UseQuadForm

Use the Quadratic Formula: $a = 1, b = - 14, c = 24$

t = \frac{- ( - 1 4 ) \pm ( - 1 4 ) ^{2} - 4 \cdot 1 \cdot 2 4}{2 \cdot 1}

NegNeg

$- (- a) = a$

t = \frac{1 4 \pm ( - 1 4 ) ^{2} - 4 \cdot 1 \cdot 2 4}{2 \cdot 1}

CalcPowProd

Calculate power and product

t = \frac{1 4 \pm 1 9 6 - 9 6}{2}

SubTerm

Subtract term

t = \frac{1 4 \pm 1 0 0}{2}

CalcRoot

Calculate root

t = \frac{1 4 \pm 1 0}{2}

The solutions for this equation are

t = \frac{1 4 \pm 1 0}{2},

which can be separated and evaluated as follows.

$t = \frac{1 4 \pm 1 0}{2}$
$t = \frac{1 4 + 1 0}{2}$	$t = \frac{1 4 - 1 0}{2}$
$t = 12$	$t = 2$

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 (\frac{1}{t} + \frac{1}{t - 6}) = 1

Substitute

$t = 12$

4 (\frac{1}{1 2} + \frac{1}{1 2 - 6}) = ? 1

▼

Evaluate left-hand side

SubTerm

Subtract term

4 (\frac{1}{1 2} + \frac{1}{6}) = ? 1

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 2}{b \cdot 2}$

4 (\frac{1}{1 2} + \frac{2}{1 2}) = ? 1

AddFrac

Add fractions

4 (\frac{3}{1 2}) = ? 1

Multiply

\frac{1 2}{1 2} = ? 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.

Discussion

Rational Inequalities and Finding Their Solution Sets

A rational inequality is an inequality that contains a rational expression.

\frac{x + 4}{x + 3} + \frac{1}{x - 5} > \frac{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.

\frac{x ^{2} - 2 x - 8}{x - 6} \geq 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.

\frac{4}{x + 1} + \frac{x + 1}{x ^{2} - 4 x - 5} \geq \frac{- 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.

\frac{4}{x + 1} + \frac{x + 1}{x ^{2} - 4 x - 5} \geq \frac{- 3}{x - 5}

The excluded values are the ones that make the denominator

0 .

Denominator	Factored Form	Excluded Values
$x + 1$	$-$	$x = - 1$
$x^{2} - 4 x - 5$	$(x + 1) (x - 5)$	$x = - 1$ 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.

\underline{Rational Inequality} \frac{4}{x + 1} + \frac{x + 1}{x ^{2} - 4 x - 5} \geq \frac{- 3}{x - 5} ⇓ \underline{Related Rational Equation} \frac{4}{x + 1} + \frac{x + 1}{x ^{2} - 4 x - 5} = \frac{- 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.

\frac{4}{x + 1} + \frac{x + 1}{x ^{2} - 4 x - 5} = \frac{- 3}{x - 5}

▼

Simplify

MultEqn

$LHS \cdot (x + 1) (x - 5) = RHS \cdot (x + 1) (x - 5)$

(\frac{4}{x + 1} + \frac{x + 1}{x ^{2} - 4 x - 5}) (x + 1) (x - 5) = (\frac{- 3}{x - 5}) (x + 1) (x - 5)

Distr

Distribute $(x + 1) (x - 5)$

\frac{4 ( x + 1 ) ( x - 5 )}{x + 1} + \frac{( x + 1 ) ( x + 1 ) ( x - 5 )}{x ^{2} - 4 x - 5} = \frac{- 3 ( x + 1 ) ( x - 5 )}{x - 5}

Substitute

$x^{2} - 4 x - 5 = (x + 1) (x - 5)$

\frac{4 ( x + 1 ) ( x - 5 )}{x + 1} + \frac{( x + 1 ) ( x + 1 ) ( x - 5 )}{( x + 1 ) ( x - 5 )} = \frac{- 3 ( x + 1 ) ( x - 5 )}{x - 5}

CancelCommonFac

Cancel out common factors

\frac{4 ( x + 1 ) ( x - 5 )}{x + 1} + \frac{( x + 1 ) ( x + 1 ) ( x - 5 )}{( x + 1 ) ( x - 5 )} = \frac{- 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

Distr

Distribute $4$

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

Distr

Distribute $- 3$

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

AddSubTerms

Add and subtract terms

5 x - 19 = - 3 x - 3

AddEqn

$LHS + 3 x = RHS + 3 x$

8 x - 19 = - 3

AddEqn

$LHS + 19 = RHS + 19$

8 x = 16

DivEqn

$LHS / 8 = RHS / 8$

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.

\frac{4}{x + 1} + \frac{x + 1}{x ^{2} - 4 x - 5} \geq \frac{- 3}{x - 5}

Substitute

$x = - 2$

\frac{4}{- 2 + 1} + \frac{- 2 + 1}{( - 2 ) ^{2} - 4 ( - 2 ) - 5} \geq ? \frac{- 3}{- 2 - 5}

▼

Evaluate left-hand side

CalcPow

Calculate power

\frac{4}{- 2 + 1} + \frac{- 2 + 1}{4 - 4 ( - 2 ) - 5} \geq ? \frac{- 3}{- 2 - 5}

MultNegNegOnePar

$- a (- b) = a \cdot b$

\frac{4}{- 2 + 1} + \frac{- 2 + 1}{4 + 8 - 5} \geq ? \frac{- 3}{- 2 - 5}

AddSubTerms

Add and subtract terms

\frac{4}{- 1} + \frac{- 1}{7} \geq ? \frac{- 3}{- 2 - 5}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 7}{b \cdot 7}$

\frac{2 8}{- 7} + \frac{- 1}{7} \geq ? \frac{- 3}{- 2 - 5}

MoveNegDenomToNum

Put minus sign in numerator

\frac{- 2 8}{7} + \frac{- 1}{7} \geq ? \frac{- 3}{- 2 - 5}

AddFrac

Add fractions

\frac{- 2 9}{7} \geq ? \frac{- 3}{- 2 - 5}

▼

Evaluate right-hand side

SubTerm

Subtract term

\frac{- 2 9}{7} \geq ? \frac{- 3}{- 7}

DivNegNeg

$\frac{- a}{- b} = \frac{a}{b}$

\frac{- 2 9}{7} ≱ \frac{3}{7} \times

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	$- \frac{2 9}{7} ≱ \frac{3}{7}$	$\frac{1 9}{5} \geq \frac{3}{5}$	$\frac{1}{2} ≱ \frac{3}{2}$	$\frac{1 1}{7} \geq 3$

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, \infty)

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.

\underline{Solution Set} (- 1, 2] \cup (5, \infty)

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) = 5 x + 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) = \frac{C ( x )}{x} .

A (x) = \frac{C ( x )}{x} ⇕ A (x) = \frac{5 x + 2 0 0 0}{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 \Leftrightarrow \frac{5 x + 2 0 0 0}{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.

\frac{5 x + 2 0 0 0}{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.

Rational Inequality \frac{5 x + 2 0 0 0}{x} < 30 Related Equation \frac{5 x + 2 0 0 0}{x} = 30

To solve the equation, multiply both sides by

x .

\frac{5 x + 2 0 0 0}{x} = 30

▼

Solve for

x

MultEqn

$LHS \cdot x = RHS \cdot x$

5 x + 2000 = 30 x

SubEqn

$LHS - 5 x = RHS - 5 x$

2000 = 25 x

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.

\frac{5 x + 2 0 0 0}{x} < 30

Substitute

$x = - 1$

\frac{5 ( - 1 ) + 2 0 0 0}{- 1} < ? 30

▼

Evaluate left-hand side

MultPosNeg

$a (- b) = - a \cdot b$

\frac{- 5 + 2 0 0 0}{- 1} < ? 30

AddTerms

Add terms

\frac{1 9 9 5}{- 1} < ? 30

MoveNegDenomToFrac

Put minus sign in front of fraction

- \frac{1 9 9 5}{1} < ? 30

DivByOne

$\frac{a}{1} = a$

- 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 .

\underline{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

24

games

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

\frac{2 4}{4 0} .

\frac{2 4}{4 0} = 0.6 \Leftrightarrow 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

\frac{2 4 + x}{4 0 + x} = 0.75

Now solve this rational equation by multiplying both sides by

40 + x

and simplifying.

\frac{2 4 + x}{4 0 + x} = 0.75

MultEqn

$LHS \cdot (40 + x) = RHS \cdot (40 + x)$

24 + x = 0.75 (40 + x)

▼

Solve for

x

Distr

Distribute $0.75$

24 + x = 30 + 0.75 x

SubEqn

$LHS - 24 = RHS - 24$

x = 6 + 0.75 x

SubEqn

$LHS - 0.75 x = RHS - 0.75 x$

0.25 x = 6

DivEqn

$LHS / 0.25 = RHS / 0.25$

x = 24

If Dominika wins the next

24

games, her wins to losses percentage will be

75 % .

Note that this solution is not extraneous.

\frac{2 4 + x}{4 0 + x} = 0.75

Substitute

$x = 24$

\frac{2 4 + 2 4}{4 0 + 2 4} = ? 0.75

▼

Evaluate left-hand side

AddTerms

Add terms

\frac{4 8}{6 4} = ? 0.75

CalcQuot

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

40 + x

should be equal to

0.90 .

\frac{2 4 + x}{4 0 + x} = 0.90

MultEqn

$LHS \cdot (40 + x) = RHS \cdot (40 + x)$

24 + x = 0.90 (40 + x)

▼

Solve for

x

Distr

Distribute $0.90$

24 + x = 36 + 0.90 x

SubEqn

$LHS - 24 = RHS - 24$

x = 12 + 0.90 x

SubEqn

$LHS - 0.90 x = RHS - 0.90 x$

0.10 x = 12

DivEqn

$LHS / 0.10 = RHS / 0.10$

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.

\frac{2 4 + x}{4 0 + x} = 0.90

Substitute

$x = 120$

\frac{2 4 + 1 2 0}{4 0 + 1 2 0} = ? 0.90

▼

Evaluate left-hand side

AddTerms

Add terms

\frac{1 4 4}{1 6 0} = ? 0.90

CalcQuot

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.

\frac{2 4 + x}{4 0 + x} = 1 ⇓ 24 + x = 40 + x ⇓ 24 = 40 \times

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.

\frac{2 0 + 1 2 0}{4 0 + 1 2 0} = \frac{1 4 4}{1 6 0}

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.

\frac{1 4 4}{1 6 0 + 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.

\frac{1 4 4}{1 6 0 + x} = 0.50

▼

Solve for

x

MultEqn

$LHS \cdot 160 + x = RHS \cdot 160 + x$

144 = 0.50 (160 + x)

Distr

Distribute $0.50$

144 = 80 + 0.50 x

SubEqn

$LHS - 80 = RHS - 80$

64 = 0.50 x

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 .

\underline{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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Catch-Up and Review

Hint

Solution

Hint

Solution

Answer

Hint

Solution

Extra

Hint

Solution

Hint

Solution

Excluded Values

Solve the Related Equation

Divide the Number Line Into Intervals

Test Values

Answer

Hint

Solution