01.6.1 Introduction to Rational Inequalities

A rational expression is a quotient of two polynomials. Its general form is:

\[ R(x)=\frac{P(x)}{Q(x)} \]

where:

  • \(P(x)\) is a polynomial called the numerator.
  • \(Q(x)\) is a polynomial called the denominator.
  • \(Q(x)\neq0\).

A rational inequality compares a rational expression with another expression using an inequality symbol. Common forms include:

\[ \frac{P(x)}{Q(x)}>0 \]
\[ \frac{P(x)}{Q(x)}<0 \]
\[ \frac{P(x)}{Q(x)}\geq0 \]
\[ \frac{P(x)}{Q(x)}\leq0 \]

Examples of rational inequalities

Example A

\[ \frac{x-3}{x+2}>0 \]

This is a rational inequality because it contains a fraction with polynomial expressions. Example B

\[ \frac{2x+1}{x-4}\leq0 \]

This is also a rational inequality. Example C

\[ \frac{x^2-9}{x+1}\geq0 \]

Here, the numerator is a quadratic polynomial, while the denominator is linear. Example D

\[ \frac{1}{x-2}>3 \]

This is another rational inequality, even though the right-hand side is not zero. Before using sign analysis, we should rewrite this inequality with zero on one side.

Rational equations vs. rational inequalities

Consider the rational equation:

\[ \frac{x-2}{x+3}=0 \]

A fraction equals zero when its numerator equals zero and its denominator is nonzero. Therefore:

\[ x-2=0 \]
\[ x=2 \]

Since \(2+3\neq0\), the equation has the solution \(x=2\). Now consider the rational inequality:

\[ \frac{x-2}{x+3}>0 \]

This requires the entire fraction to be positive. That can happen when:

  • Both numerator and denominator are positive.
  • Both numerator and denominator are negative.

Its solution is:

\[ x<-3\quad\text{or}\quad x>2 \]

We will study the complete solving process later in this lesson. Key concept: Rational inequalities usually produce intervals of solutions rather than just individual values.

 

01.6.2 Understanding Domain Restrictions

The first essential rule when working with rational expressions is: A denominator cannot equal zero. For example:

\[ \frac{5}{0} \]

is undefined. Therefore, whenever a variable appears in a denominator, we must identify the values that make the denominator zero. These values are excluded from the domain.

Example 1: Finding a domain restriction

Consider:

\[ \frac{x+4}{x-3} \]

The denominator is:

\[ x-3 \]

Set the denominator equal to zero:

\[ x-3=0 \]

Therefore:

\[ x=3 \]

The expression is undefined when \(x=3\). Domain restriction:

\[ x\neq3 \]

The domain is:

\[ (-\infty,3)\cup(3,\infty) \]

Example 2: Multiple restrictions

Consider:

\[ \frac{x+2}{(x-1)(x+4)} \]

The denominator must not equal zero:

\[ (x-1)(x+4)\neq0 \]

The denominator becomes zero when:

x − 1 = 0 ⟹ x = 1

or:

x + 4 = 0 ⟹ x = −4

 

 

Therefore:

\[ x\neq1,\qquad x\neq-4 \]

Domain:

\[ (-\infty,-4)\cup(-4,1)\cup(1,\infty) \]

Important distinction: Zeros and undefined values

Consider:

\[ R(x)=\frac{x-2}{x+3} \]

There are two important values.

Value Reason Meaning
\(x=2\) Numerator equals zero Expression equals zero
\(x=-3\) Denominator equals zero Expression is undefined

For \(x=2\):

\[ R(2)=\frac{2-2}{2+3}=0 \]

For \(x=-3\):

\[ R(-3)=\frac{-3-2}{-3+3} =\frac{-5}{0} \]

This is undefined. Essential rule:

  • Numerator zeros may be included when the inequality contains \(\geq\) or \(\leq\), provided the denominator is nonzero.
  • Denominator zeros are never included in the solution set, regardless of the inequality symbol.
 

01.6.3 Understanding the Sign of a Fraction

To solve rational inequalities, students must understand how positive and negative numbers behave under division. Consider:

\[ \frac{A}{B} \]

where \(B\neq0\). The sign of the fraction depends on the signs of \(A\) and \(B\).

Sign Rules for Fractions

Numerator Denominator Result
Positive (+) Positive (+) Positive (+)
Negative (−) Negative (−) Positive (+)
Positive (+) Negative (−) Negative (−)
Negative (−) Positive (+) Negative (−)
Zero Nonzero Zero
Any value Zero Undefined

Example 3: Positive fractions

Consider:

\[ \frac{6}{3}=2>0 \]

Both numerator and denominator are positive. Now consider:

\[ \frac{-6}{-3}=2>0 \]

Both numerator and denominator are negative. In both cases, the fraction is positive.

Example 4: Negative fractions

Consider:

\[ \frac{-6}{3}=-2<0 \]

The numerator is negative, while the denominator is positive. Similarly:

\[ \frac{6}{-3}=-2<0 \]

The numerator is positive, while the denominator is negative. In both cases, the fraction is negative.

Mathematical summary

For \(B\neq0\):

\[ \frac{A}{B}>0 \]

when \(A\) and \(B\) have the same sign. And:

\[ \frac{A}{B}<0 \]

when \(A\) and \(B\) have opposite signs. This rule forms the foundation of the sign-chart method.

 

01.6.4 Critical Values and Sign Charts

A critical value in rational inequality solving is a real value where the numerator or denominator becomes zero. Critical values divide the number line into intervals. Within each interval, a rational expression is continuous and cannot change sign unless it crosses a numerator zero or encounters a denominator zero. Therefore, we can determine the sign of the expression by testing one value in each interval.

General method for solving rational inequalities

  1. Move all terms to one side so that the other side is zero.
  2. Combine the expression into a single fraction when possible.
  3. Identify values that make the original denominator zero.
  4. Factor the numerator and denominator where possible.
  5. Find all real critical values.
  6. Arrange the critical values from smallest to largest.
  7. Divide the number line into intervals.
  8. Test the sign of the expression in each interval.
  9. Select intervals satisfying the inequality.
  10. Include permitted numerator zeros for non-strict inequalities.
  11. Exclude every original denominator zero.
  12. Write the answer in interval notation.

Example 5: A basic rational inequality

Solve:

\[ \frac{x-2}{x+3}>0 \]

Step 1. Identify the numerator zero. Set:

\[ x-2=0 \]

Therefore:

\[ x=2 \]

Step 2. Identify the denominator zero. Set:

\[ x+3=0 \]

Therefore:

\[ x=-3 \]

The value \(x=-3\) is excluded from the domain. Step 3. Arrange the critical values.

\[ -3<2 \]

These values divide the number line into three intervals:

\[ (-\infty,-3) \]
\[ (-3,2) \]
\[ (2,\infty) \]

Step 4. Test each interval. For \(x<-3\), choose \(x=-4\).

\[ \frac{-4-2}{-4+3} =\frac{-6}{-1} =6>0 \]

The expression is positive. For \(-3<x<2\), choose \(x=0\).

\[ \frac{0-2}{0+3} =-\frac23<0 \]

The expression is negative. For \(x>2\), choose \(x=3\).

\[ \frac{3-2}{3+3} =\frac16>0 \]

The expression is positive.

Step 5. Construct a Sign Chart.

Interval Test value Numerator Denominator Fraction sign
(−∞, −3) −4 − − Positive (+)
(−3, 2) 0 − + Negative (−)
(2, ∞) 3 + + Positive (+)

Step 6. Choose the correct intervals. We need:

\[ \frac{x-2}{x+3}>0 \]

Therefore, select the intervals where the fraction is positive. The solution is:

\[ x<-3\quad\text{or}\quad x>2 \]

Final answer in interval notation:

\[ (-\infty,-3)\cup(2,\infty) \]

Neither critical value is included. The value −3 is excluded because the denominator equals zero, and the value 2 is excluded because the fraction equals zero while the inequality is strict.

Visual Interpretation

Solution of   x − 2 x + 3   > 0

 

 

 

 

 

 

 

+ − + −3 2

The highlighted regions represent the solution. Open circles indicate that both −3 and 2 are excluded.

01.6.5 Solving Non-Strict Rational Inequalities

A non-strict inequality uses one of the following symbols:

\[ \geq\qquad\leq \]

These symbols allow equality. For rational expressions, equality occurs when the numerator equals zero while the denominator remains nonzero.

Example 6: Rational inequality with \(\leq\)

Solve:

\[ \frac{x-4}{x+1}\leq0 \]

Step 1. Find the critical values.

Numerator:

x − 4 = 0 ⟹ x = 4

Denominator:

x + 1 = 0 ⟹ x = −1

Domain restriction:

\[ x\neq-1 \]

Step 2. Identify the intervals.

\[ (-\infty,-1),\quad(-1,4),\quad(4,\infty) \]

Step 3. Test the signs. For \(x=-2\):

\[ \frac{-2-4}{-2+1} =\frac{-6}{-1}=6>0 \]

For \(x=0\):

\[ \frac{0-4}{0+1}=-4<0 \]

For \(x=5\):

\[ \frac{5-4}{5+1}=\frac16>0 \]

Step 4. Determine the solution. We require the fraction to be negative or zero. The fraction is negative between −1 and 4. At \(x=4\), the numerator equals zero and the denominator is nonzero, so the value is included. At \(x=-1\), the expression is undefined, so the value is excluded. The solution is:

\[ -1<x\leq4 \]

Final answer:

\[ (-1,4] \]

Strict and non-strict inequality comparison

For the expression x − 4 x + 1 :

Inequality Solution set
  x − 4 x + 1 > 0 (−∞, −1) ∪ (4, ∞)
  x − 4 x + 1 ≥ 0 (−∞, −1) ∪ [4, ∞)
  x − 4 x + 1 < 0 (−1, 4)
  x − 4 x + 1 ≤ 0 (−1, 4]

Important: Notice that x = −1 never appears as an included endpoint because the denominator becomes zero at this value.

01.6.6 Rational Inequalities with Quadratic Numerators

Some rational inequalities contain quadratic expressions in the numerator. In these cases, we can apply the factorization techniques learned in Lesson 01.5.

Example 7: Quadratic numerator

Solve:

\[ \frac{x^2-x-2}{x-3}\geq0 \]

Step 1. Factorize the numerator.

\[ x^2-x-2=(x-2)(x+1) \]

Therefore:

\[ \frac{(x-2)(x+1)}{x-3}\geq0 \]

Step 2. Find the critical values. Numerator zeros:

\[ x=2,\quad x=-1 \]

Denominator zero:

\[ x=3 \]

Domain restriction:

\[ x\neq3 \]

Step 3. Arrange the critical values.

\[ -1<2<3 \]
 

The intervals are:

(−∞, −1), (−1, 2), (2, 3), (3, ∞)

Step 4. Determine the signs.

Interval Test value Sign of numerator Sign of denominator Fraction sign
(−∞, −1) −2 + − −
(−1, 2) 0 − − +
(2, 3) 2.5 + − −
(3, ∞) 4 + + +

Step 5. Select the intervals.

The inequality requires:

    (x − 2)(x + 1) x − 3   ≥ 0

Therefore, select the positive intervals and any permitted numerator zeros. The positive intervals are:

\[ (-1,2) \]

and:

\[ (3,\infty) \]

The numerator equals zero at \(x=-1\) and \(x=2\), both of which are allowed. The denominator equals zero at \(x=3\), so this value must be excluded. Final answer:

\[ [-1,2]\cup(3,\infty) \]
 

01.6.7 Why We Cannot Multiply by an Unknown Denominator

This is one of the most important concepts in the entire lesson. In Lesson 01.4, we established:

  • Multiplying an inequality by a positive number preserves its direction.
  • Multiplying an inequality by a negative number reverses its direction.

Now consider:

\[ \frac{1}{x-2}>1 \]

We might try to multiply both sides by \(x-2\). However, there is a problem. We do not yet know whether:

\[ x-2>0 \]

or:

\[ x-2<0 \]

Therefore, we cannot determine whether the inequality sign should remain unchanged or reverse. Incorrect shortcut: Multiplying by \(x-2\) while automatically keeping the sign:

\[ 1>x-2 \]

This is not a valid equivalent step unless we first establish that \(x-2\) is positive.

Correct method

Solve:

\[ \frac{1}{x-2}>1 \]

Step 1. Move all terms to the left.

\[ \frac{1}{x-2}-1>0 \]

Step 2. Combine into a single fraction.

\[ \frac{1-(x-2)}{x-2}>0 \]

Simplify:

\[ \frac{3-x}{x-2}>0 \]

Step 3. Find the critical values.

Numerator zero:

3 − x = 0 ⟹ x = 3

Denominator zero:

x − 2 = 0 ⟹ x = 2

Step 4. Determine the Signs.

Interval Numerator Denominator Fraction
x < 2 Positive Negative Negative
2 < x < 3 Positive Positive Positive
x > 3 Negative Positive Negative

Step 5. Select the positive interval.

\[ 2<x<3 \]

Final answer:

\[ (2,3) \]

The inequality is satisfied only for values strictly between 2 and 3. Important: You may multiply by a variable expression if its sign is established or if you correctly analyze separate sign cases. For ordinary rational inequalities, combining terms and constructing a sign chart is usually safer and more efficient.

 

01.6.8 Solving Inequalities with Multiple Rational Expressions

Sometimes a rational inequality contains fractions on both sides. In these situations, we usually need to combine the expressions into a single fraction before performing sign analysis.

Example 8: Rational inequality with fractions on both sides

Solve:

\[ \frac{1}{x-1}\geq\frac{2}{x+2} \]

Step 1. Identify the domain restrictions. The denominators must not equal zero.

x − 1 ≠ 0  ⟹   x ≠ 1
x + 2 ≠ 0  ⟹  x ≠ −2

Step 2. Move everything to the left.

\[ \frac{1}{x-1}-\frac{2}{x+2}\geq0 \]

Step 3. Find the common denominator. The common denominator is:

\[ (x-1)(x+2) \]

Combine the fractions:

\[ \frac{(x+2)-2(x-1)}{(x-1)(x+2)}\geq0 \]

Expand the numerator:

\[ \frac{x+2-2x+2}{(x-1)(x+2)}\geq0 \]

Simplify:

\[ \frac{4-x}{(x-1)(x+2)}\geq0 \]

Step 4. Find the critical values. Numerator:

4 − x = 0

⟹

x = 4

Denominator:

\[ x=1,\quad x=-2 \]

Arrange in ascending order:

\[ -2<1<4 \]

Step 5. Analyze the Signs.

Interval Test value Sign of fraction
(−∞, −2) −3 Positive
(−2, 1) 0 Negative
(1, 4) 2 Positive
(4, ∞) 5 Negative

Step 6. Select the required intervals. The inequality requires the fraction to be nonnegative. The positive intervals are:

\[ (-\infty,-2)\quad\text{and}\quad(1,4) \]

At \(x=4\), the numerator equals zero and the denominator is nonzero, so 4 is included. The values −2 and 1 must be excluded because they make the original denominators zero. Final answer:

\[ (-\infty,-2)\cup(1,4] \]

Important rule for combining fractions

When combining rational expressions, do not multiply both sides by an expression whose sign is unknown. Instead, create a common denominator and analyze the resulting fraction.

 

01.6.9 Rational Inequalities with Quadratic Denominators

Rational inequalities can also contain quadratic polynomials in the denominator. In such problems, factorizing the denominator often produces several critical values.

Example 9: Quadratic denominator

Solve:

\[ \frac{x-2}{x^2+2x-3}<0 \]

Step 1. Factorize the denominator.

\[ x^2+2x-3=(x+3)(x-1) \]

Therefore:

\[ \frac{x-2}{(x+3)(x-1)}<0 \]

Step 2. Find the critical values.

Numerator:

x − 2 = 0

⟹

x = 2

Denominator:

x + 3 = 0

⟹

x = −3

x − 1 = 0

⟹

x = 1

The excluded domain values are:

\[ x\neq-3,\quad x\neq1 \]

Step 3. Arrange the critical values.

\[ -3<1<2 \]

Step 4. Determine the Signs.

Interval x − 2 x + 3 x − 1 Overall sign
x < −3 − − − Negative
−3 < x < 1 − + − Positive
1 < x < 2 − + + Negative
x > 2 + + + Positive
 

Step 5. Select the negative intervals. We need:

\[ \frac{x-2}{(x+3)(x-1)}<0 \]

Therefore, select the regions with a negative sign. Final answer:

\[ (-\infty,-3)\cup(1,2) \]

All three critical values are excluded: −3 and 1 make the denominator zero, while 2 makes the fraction zero rather than negative.

 

01.6.10 Canceling Common Factors and Removable Discontinuities

Sometimes a rational expression contains the same factor in both the numerator and denominator. We can cancel common factors, but there is an important restriction. Canceling a factor does not make its original denominator zero permissible.

Example 10: A canceled factor

Solve:

\[ \frac{x^2-1}{x-1}\geq0 \]

Step 1. Identify the original domain restriction.

\[ x-1\neq0 \]

Therefore:

\[ x\neq1 \]

Step 2. Factorize the numerator.

\[ x^2-1=(x-1)(x+1) \]

Rewrite:

\[ \frac{(x-1)(x+1)}{x-1}\geq0 \]

Step 3. Cancel the common factor. For \(x\neq1\):

\[ x+1\geq0 \]

Step 4. Solve the linear inequality.

\[ x\geq-1 \]

Step 5. Apply the original domain restriction. The simplified inequality suggests all values greater than or equal to −1. However, \(x=1\) must still be excluded because the original expression is undefined there. Final answer:

\[ [-1,1)\cup(1,\infty) \]

Why is \(x=1\) excluded?

Substituting \(x=1\) into the original fraction gives:

\[ \frac{1^2-1}{1-1}=\frac00 \]

This expression is undefined. The simplified expression \(x+1\) equals 2 at \(x=1\), but that does not change the domain of the original problem. This missing point is called a removable discontinuity, also known as a hole in the graph.

Key Distinction

Situation Meaning
Factor cancels algebraically The expression simplifies on its original domain
Original denominator equals zero The value remains excluded
Simplified expression is defined at the excluded value The original expression is still undefined there

This is one of the most common sources of incorrect answers in rational inequalities.

01.6.11 Repeated Factors and Sign Changes

Some rational inequalities contain repeated factors, such as:

\[ (x-2)^2 \]

A squared factor is always nonnegative. Therefore, the sign behavior around a repeated factor differs from the behavior around a factor raised to an odd power.

Example 11: A repeated factor in the numerator

Solve:

\[ \frac{(x-2)^2}{x+1}>0 \]

Step 1. Find the critical values. Numerator zero:

\[ x=2 \]

Denominator zero:

\[ x=-1 \]

Domain restriction:

\[ x\neq-1 \]

Step 2. Analyze the numerator. Since:

\[ (x-2)^2\geq0 \]

the numerator is positive for every real \(x\neq2\). At \(x=2\), the numerator equals zero. Step 3. Analyze the denominator.

\[ x+1<0\quad\text{when }x<-1 \]
\[ x+1>0\quad\text{when }x>-1 \]

Step 4. Construct a Sign Chart.

Interval Numerator Denominator Fraction
x < −1 Positive Negative Negative
−1 < x < 2 Positive Positive Positive
x = 2 Zero Positive Zero
x > 2 Positive Positive Positive

Step 5. Select the positive values. We need a fraction strictly greater than zero. Therefore, \(x=2\) must be excluded because the fraction equals zero. The denominator zero at \(x=-1\) is also excluded. Final answer:

\[ (-1,2)\cup(2,\infty) \]

Important observation

The expression is positive immediately before and after \(x=2\). It does not change sign when crossing this repeated numerator zero.

Even and Odd Factor Powers

For a factor (x − a)m :

Power m Sign behavior across x = a
Odd Factor changes sign
Even Factor does not change sign

A denominator factor of even multiplicity also preserves the sign across its zero, but the point itself remains undefined. This principle is useful for solving rational inequalities containing several factors.

 

01.6.12 Special Cases: No Solution, One Solution, or Almost All Real Numbers

A rational inequality does not always produce one or two ordinary intervals. Some rational expressions have predictable signs because their numerators and denominators contain squares.

Case 1: No real solution

Solve:

\[ \frac{x^2+4}{x^2+1}<0 \]

Observe:

\[ x^2+4>0 \]

and:

\[ x^2+1>0 \]

for every real \(x\). Therefore, the numerator and denominator are always positive. The entire fraction is always positive. It can never be less than zero.

Final answer:

∅

Case 2: Exactly one solution

Solve:

\[ \frac{(x-1)^2}{x^2+1}\leq0 \]

The denominator is always positive:

\[ x^2+1>0 \]

The numerator is always nonnegative:

\[ (x-1)^2\geq0 \]

Therefore, the fraction cannot be negative. To satisfy the inequality, the fraction must equal zero. Set:

\[ (x-1)^2=0 \]

Therefore:

\[ x=1 \]

Final answer:

\[ \{1\} \]

Case 3: All real numbers except an undefined point

Solve:

\[ \frac{x^2+1}{(x-2)^2}\geq0 \]

The numerator is always positive. The denominator is positive whenever \(x\neq2\). Therefore, the fraction is positive for every value in its domain. However, \(x=2\) must be excluded. Final answer:

\[ (-\infty,2)\cup(2,\infty) \]

Summary of Special Cases

Type Possible solution set
Expression never meets the inequality Empty set
Equality occurs at one permitted value Singleton set
Inequality holds everywhere in its domain Entire domain
Expression changes sign One or more intervals

01.6.13 A Useful Alternative: Multiplying by the Square of the Denominator

There is another mathematically valid method for solving rational inequalities. Consider:

\[ \frac{P(x)}{Q(x)}>0 \]

where:

\[ Q(x)\neq0 \]

We know that:

\[ [Q(x)]^2>0 \]

for every value in the domain. Therefore, we may multiply both sides by \([Q(x)]^2\) without reversing the inequality sign. This gives:

\[ P(x)Q(x)>0 \]

provided that \(Q(x)\neq0\) remains an explicit restriction.

Example 12: The squared-denominator method

Solve:

\[ \frac{x-1}{x+2}\geq0 \]

Step 1. Identify the restriction.

\[ x+2\neq0 \]

Therefore:

\[ x\neq-2 \]

Step 2. Multiply both sides by \((x+2)^2\). Since this square is positive for all allowed values of \(x\), the inequality direction remains unchanged.

\[ (x-1)(x+2)\geq0 \]

Step 3. Solve the resulting quadratic inequality. The roots are:

\[ x=1,\quad x=-2 \]

The quadratic expression is nonnegative outside these roots. Therefore:

\[ x\leq-2\quad\text{or}\quad x\geq1 \]

Step 4. Apply the domain restriction. Since \(x=-2\) makes the original denominator zero, it must be excluded. Final answer:

\[ (-\infty,-2)\cup[1,\infty) \]

This method works because we multiply by a quantity known to be positive. It can be useful for transforming rational inequalities into polynomial inequalities. However, students must remember the original domain restrictions. Recommended exam approach: Use the sign-chart method as your primary technique. Use the squared-denominator method when it makes the calculation simpler.

 

01.6.14 Graphical Interpretation of Rational Inequalities

A rational inequality can also be understood through the graph of a rational function. Consider:

\[ y=\frac{x-2}{x+3} \]

The graph has three important features.

1. X-intercept

The graph crosses the x-axis when:

\[ y=0 \]

This happens when:

\[ x-2=0 \]

Therefore:

\[ x=2 \]

The graph has an x-intercept at:

\[ (2,0) \]

2. Undefined point

The denominator equals zero when:

\[ x+3=0 \]

Therefore:

\[ x=-3 \]

This rational function has a vertical asymptote at \(x=-3\). The graph is not defined at that value. For rational expressions containing canceled factors, an excluded denominator zero may instead correspond to a removable discontinuity.

3. Positive and negative regions

The inequality:

\[ \frac{x-2}{x+3}>0 \]

means:

\[ y>0 \]

Therefore, we need the regions where the graph lies above the x-axis. Similarly:

\[ \frac{x-2}{x+3}<0 \]

means that the graph lies below the x-axis.

Graphical Interpretation Summary

Inequality Graphical meaning
R(x) > 0 Graph above the x-axis
R(x) < 0 Graph below the x-axis
R(x) ≥ 0 Graph above or on the x-axis
R(x) ≤ 0 Graph below or on the x-axis

Undefined points are always excluded, even if the inequality includes equality.

01.6.15 Applying Rational Inequalities to Word Problems

Rational inequalities are useful in problems involving speed, time, production, cost, efficiency, and other quantities expressed as ratios.

Example 13: Travel time problem

A student needs to travel 120 kilometers to an examination center. The journey must take less than 3 hours. Assuming constant speed, what is the minimum speed requirement? Step 1. Define the variable. Let \(v\) represent the speed in kilometers per hour. Since speed is positive:

\[ v>0 \]

Step 2. Write the time formula.

Step 2. Write the Time Formula.

Recall:

  Time =Distance        Speed

Therefore:

\[ t=\frac{120}{v} \]

Step 3. Construct the inequality. The journey must take less than 3 hours.

\[ \frac{120}{v}<3 \]

Step 4. Solve the inequality. Since \(v>0\), we may multiply both sides by \(v\) without reversing the inequality sign.

\[ 120<3v \]

Divide by 3:

\[ 40<v \]

Therefore:

\[ v>40 \]

Final answer: The student must travel at an average speed greater than 40 km/h. At exactly 40 km/h, the journey would take exactly 3 hours, which does not satisfy the strict time requirement.

Why was multiplication by \(v\) allowed?

Because the problem explicitly states:

\[ v>0 \]

We already know that \(v\) is positive. Therefore, the inequality direction remains unchanged.

 

01.6.16 Common Mistakes When Solving Rational Inequalities

Mistake 1: Including a denominator zero

Consider:

\[ \frac{x-1}{x+2}\geq0 \]

An incorrect answer is:

\[ (-\infty,-2]\cup[1,\infty) \]

The correct answer is:

\[ (-\infty,-2)\cup[1,\infty) \]

The value −2 must be excluded because the denominator equals zero.

Mistake 2: Multiplying by an expression with an unknown sign

Consider:

\[ \frac{1}{x-3}>2 \]

Multiplying both sides by \(x-3\) without checking its sign is not valid. The correct approach is to write:

\[ \frac{1-2(x-3)}{x-3}>0 \]

and analyze the sign of the resulting fraction.

Mistake 3: Canceling a factor and forgetting the restriction

Consider:

\[ \frac{(x-1)(x+3)}{x-1}>0 \]

The expression simplifies to:

\[ x+3>0 \]

but only when:

\[ x\neq1 \]

The correct solution is:

\[ (-3,1)\cup(1,\infty) \]

Mistake 4: Assuming the sign always alternates

If a rational expression contains a repeated factor, the sign may not change when crossing that factor’s zero. For example:

\[ (x-2)^2 \]

is positive on both sides of \(x=2\), and equals zero at \(x=2\). Therefore, do not automatically alternate signs without examining multiplicities or testing intervals.

Mistake 5: Finding only the numerator zeros

For a rational inequality, finding the numerator zeros is not enough. Students must also identify every denominator zero and determine where the fraction has the required sign.

Mistake 6: Forgetting the original domain after simplifying

A simplified rational expression may be defined at values where the original expression was not defined. Always preserve the original domain restrictions.

Mistake 7: Confusing AND with OR

When the solution consists of two separate intervals, we use the union symbol:

\[ \cup \]

For example:

\[ (-\infty,-3)\cup(2,\infty) \]

When a problem requires two conditions simultaneously, we use their intersection.

01.6.17 General Strategy for Solving Rational Inequalities

When solving any rational inequality, use the following systematic process.

01.6.17 General Strategy for Solving Rational Inequalities

When solving any rational inequality, use the following systematic process.

Step 1: Write the inequality in standard form
Move everything to one side so the other side is zero.
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Step 2: Identify the original domain restrictions
Find every value that makes an original denominator zero.
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Step 3: Combine and factor the expression
Write a single rational expression and factor where possible.
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Step 4: Find the critical values
Identify numerator zeros and denominator zeros, including canceled factors.
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Step 5: Construct a sign chart
Arrange the critical values and determine signs in each interval.
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Step 6: Select the solution intervals
Choose the regions that satisfy the original inequality.
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Step 7: Check excluded and included points
Include permitted numerator zeros for non-strict inequalities. Never include denominator zeros.
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Step 8: Write and verify the answer
Use interval notation and check selected test values.

Choosing the Most Efficient Solving Method

Situation Recommended approach
One linear factor in numerator and denominator Basic sign chart
Multiple factors Complete factor sign table
Quadratic numerator or denominator Factorize, then use sign analysis
Fractions on both sides Combine into one fraction
Common factors cancel Simplify but preserve original restrictions
Repeated factors Analyze multiplicity and sign behavior
Expression has predictable sign Use square properties
Domain restrictions have known signs Use valid inequality multiplication rules

01.6.18 Lesson Summary

This lesson introduced rational inequalities and developed a systematic approach to solving them. Students should remember the following essential principles:

  1. Rational inequalities compare rational expressions using \(>,<,\geq,\leq\).
  2. Every value that makes an original denominator zero must be excluded.
  3. Numerator zeros may be included only when the inequality permits equality and the denominator is nonzero.
  4. A rational expression is positive when its numerator and denominator have the same sign.
  5. A rational expression is negative when its numerator and denominator have opposite signs.
  6. Critical values divide the number line into intervals for sign analysis.
  7. One test value per interval is sufficient once all critical values have been identified.
  8. Multiplying by an expression of unknown sign is not valid without considering the sign cases.
  9. Canceling common factors does not remove original domain restrictions.
  10. Repeated even-power factors do not change sign across their zeros.
  11. The final answer should express the entire solution set, not merely critical values.
  12. Interval notation must accurately show included and excluded endpoints.

Essential rules to memorize

For \(B\neq0\):

\[ \frac{A}{B}>0 \]

when \(A\) and \(B\) have the same sign.

\[ \frac{A}{B}<0 \]

when \(A\) and \(B\) have opposite signs. And:

\[ \frac{A}{B}=0 \]

when \(A=0\) and \(B\neq0\). Final reminder: In rational inequalities, finding the zeros, determining the signs, and checking domain restrictions are three separate tasks. All three are necessary to obtain the correct solution.

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