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} -

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Catch-Up and Review

Hint

Solution

Hint

Solution

Answer

Hint

Solution