01.7.1 Essential Review: Sets and Inequality Notation

Before attempting mixed problems, students should review the mathematical symbols introduced in earlier lessons.

Set notation

A set is a well-defined collection of distinct objects or elements.

For example:

The numbers 1, 2, 3, 4, and 5 are the elements of set \(A\).

If an element belongs to a set, we write:

If an element does not belong to a set, we write:

Main set operations

Suppose:

The union of \(A\) and \(B\) contains every element appearing in at least one of the sets.

The intersection contains elements common to both sets.

The difference \(A\setminus B\) contains elements belonging to \(A\) but not \(B\).

The complement \(A^c\) contains the elements of a specified universal set \(U\) that do not belong to \(A\).

For example, if:

Then:

The complement depends on the universal set. Without knowing the relevant universe, it may be impossible to determine a complement exactly.

01.7.2 Connecting Set Operations with Inequalities

A central idea of this lesson is that the solution to an inequality can itself be treated as a set.

For example:

defines the solution set:

And:

defines:

We can use set operations to combine these inequalities.

Case 1: AND means intersection

Suppose we require:

Both conditions must hold simultaneously.

Therefore:

Equivalent inequality notation:

Case 2: OR means union

Now suppose we require:

A number satisfies the condition if it belongs to at least one of the solution sets.

Since every real number satisfies at least one of these two conditions:

Important: The words AND and OR describe different mathematical operations. Confusing them may produce an entirely different answer.

AND vs. OR reference table

01.7.3 Worked Example 1 — Mixed Set Operations

Let the universal set be:

Let:

Find:

1. \(A\cap B\)

2. \(A\cup B\)

3. \(A\setminus B\)

4. \((A\cup B)^c\)

Solution

Part 1. Intersection

Find the elements appearing in both sets.

The common elements are 2, 5, and 8.

Therefore:

Part 2. Union

Collect every distinct element belonging to either set.

Remember that repeated elements are written only once in a set.

Part 3. Set difference

We need elements belonging to \(A\) but not \(B\).

Notice that \(A\setminus B\) is not generally the same as \(B\setminus A\).

Part 4. Complement of the union

First identify:

Now find the elements in \(U\) that do not appear in this union.

Exam strategy

For problems involving several set operations, work from the innermost parentheses outward.

For example, in:

calculate \(A\cup B\) first, then take its complement relative to \(U\).

01.7.4 Worked Example 2 — Venn Diagram and Counting

A class contains 40 students.

Among them:

• 24 students study mathematics.

• 19 students study physics.

• 8 students study both mathematics and physics.

Find the number of students who study at least one of the two subjects and the number who study neither subject.

Step 1. Define the sets

Let:

We are given:

Step 2. Apply the inclusion-exclusion principle

For two finite sets:

Substituting:

Therefore, 35 students study at least one of the two subjects.

Why do we subtract 8?

Because the students who study both subjects were counted once in the mathematics group and once in the physics group.

Without subtracting the overlap, they would be counted twice.

Step 3. Find the students who study neither subject

Total number of students:

Number studying at least one subject:

Therefore:

Final answer:

• Students studying at least one subject: 35

• Students studying neither subject: 5

Check using the four Venn regions

Verification:

The total matches the original class size.

01.7.5 Worked Example 3 — Intervals and Set Operations

Let:

Find:

Part 1. Find the intersection

The intersection contains values belonging to both intervals.

The first interval contains all values at most 5.

The second interval contains all values strictly between 2 and 9.

Therefore, the common region is:

The value 2 is excluded because it is not in \(B\), while 5 is included because it belongs to both intervals.

Part 2. Find the union

The union contains all values belonging to either interval.

Together, these intervals cover every real number less than 9.

Therefore:

The value 9 is excluded.

Part 3. Find the set difference

We need:

These are the values that belong to \(A\) but not \(B\).

The second interval removes all values greater than 2 and less than or equal to 5 from \(A\).

However, \(x=2\) is not in \(B\), so 2 remains in the difference.

Therefore:

Key lesson: When combining intervals, always check whether their endpoints are included or excluded.

01.7.6 Worked Example 4 — Compound Linear Inequalities

Solve:

Express the solution in interval notation.

Step 1. Subtract 1 from all three parts

Simplify:

Step 2. Divide all three parts by 2

Since 2 is positive, the inequality directions remain unchanged.

Step 3. Write the solution in interval notation

Step 4. Verify the endpoints

At \(x=-2\):

This satisfies the original non-strict lower bound, so −2 is included.

At \(x=4\):

But the original upper bound requires a value strictly less than 9.

Therefore, 4 must be excluded.

Final answer:

01.7.7 Worked Example 5 — Intersection of Two Linear Inequalities

Solve the system:

A system of inequalities requires all given conditions to hold simultaneously.

Step 1. Solve the first inequality

Add 5:

Divide by 3:

Its solution set is:

Step 2. Solve the second inequality

Subtract 1:

Divide by 2:

Its solution set is:

Step 3. Find the intersection

Both conditions must be satisfied.

Therefore:

Final answer:

Important: Do not write the union of these solution sets. Because both inequalities must hold, we use the intersection.

01.7.8 Worked Example 6 — Quadratic Inequality with an Interval Restriction

Find all real values satisfying:

and:

Step 1. Factorize the quadratic expression

Therefore:

Step 2. Find the roots

Step 3. Solve the quadratic inequality

The quadratic opens upward.

Therefore, it is nonpositive between its roots, including both roots.

The solution set is:

Step 4. Apply the additional restriction

The second condition is:

Thus:

We need the intersection:

Therefore:

Final answer:

Why is this a mixed problem?

It requires three separate skills:

• Solving a quadratic inequality.

• Expressing its solution using interval notation.

• Finding the intersection of two sets.

Students must complete all three operations to obtain the correct answer.

01.7.9 Worked Example 7 — Rational Inequality with an Interval Restriction

Find the values satisfying:

subject to:

Step 1. Identify the critical values

The value \(x=-2\) is excluded because the denominator cannot equal zero.

Step 2. Determine the signs

The critical values divide the number line into three intervals.

Step 3. Solve the rational inequality

The fraction is nonnegative when the numerator and denominator have the same sign, or the numerator is zero and the denominator is nonzero.

Therefore:

Solution set:

Step 4. Apply the interval restriction

We also require:

Find:

The left portion becomes:

The right portion becomes:

Combine these two regions.

Final answer:

Critical exam reminder

Although the original rational inequality permits all values less than −2, the additional restriction excludes values less than or equal to −3.

The final answer must satisfy every stated condition.

01.7.10 Worked Example 8 — A Canceled Factor and Its Original Domain

Solve:

Step 1. Identify the original denominator restriction

Therefore:

Step 2. Simplify the expression

For \(x\neq2\), cancel the common factor:

The inequality becomes:

Step 3. Solve

Step 4. Apply the original restriction

The value \(x=2\) must remain excluded because the original expression is undefined there.

Therefore:

Final answer:

This is an important example because a student who forgets the original denominator restriction might incorrectly answer [-1,\infty)” data-math-display=”false”>\([-1,\infty)\).

01.7.11 Worked Example 9 — Combining Quadratic and Rational Inequalities

Solve the following system:

This problem combines two different inequality-solving techniques.

Step 1. Solve the quadratic inequality

Factorize:

The roots are:

The quadratic opens upward, so its value is nonnegative outside the roots.

Therefore:

Solution set:

Step 2. Solve the rational inequality

Critical values:

The numerator and denominator have opposite signs between −1 and 4.

Therefore:

Solution set:

Step 3. Find the intersection

We need:

Substitute:

The first branch \(x\leq-2\) does not overlap \((-1,4)\).

The second branch \(x\geq3\) overlaps it from 3 to 4.

Therefore:

Final answer:

Verification

At \(x=3\):

And:

Thus, 3 is included.

At \(x=4\), the rational expression is undefined, so 4 is excluded.

01.7.12 Worked Example 10 — Integer Solution Sets

Find all integers satisfying:

and:

Step 1. Solve the quadratic inequality

Factorize:

Therefore:

Step 2. Apply the interval restriction

The additional condition is:

Intersecting the solution sets gives:

Step 3. Apply the integer restriction

The problem asks for integers only.

The integers in these intervals are:

Final answer:

Important distinction

If the problem asks for real solutions, the interval answer is correct.

If it asks for integer solutions, the answer must contain only integers satisfying all conditions.

Always check the required number domain:

The notation \(\mathbb N\) is not universal: some textbooks include 0, while others begin at 1. Follow the convention stated in the question.

01.7.13 Worked Example 11 — Recognizing an Empty Intersection

Solve both conditions:

and:

Step 1. Solve the quadratic inequality

The expression is negative between its roots.

Therefore:

Solution set:

Step 2. Express the second condition

Step 3. Find the intersection

There is no overlap between the two intervals.

Therefore:

Final answer: No real solution.

Important: A mixed inequality problem may have no solution even when each individual inequality has infinitely many solutions.

01.7.14 Worked Example 12 — Word Problem with an Integer Restriction

A student has a maximum budget of $60 for examination preparation materials.

The student must pay a fixed registration fee of $12. Each additional practice workbook costs $7.

Find the possible numbers of workbooks the student can buy without exceeding the budget.

Step 1. Define the variable

Let \(n\) be the number of workbooks.

Since the number of workbooks must be a nonnegative integer:

Step 2. Construct the inequality

Fixed registration fee:

Cost of \(n\) workbooks:

Total cost cannot exceed $60.

Therefore:

Step 3. Solve

Subtract 12:

Divide by 7:

Since:

the maximum possible integer value is 6.

Step 4. Express the final solution set

Final answer: The student may buy up to 6 workbooks.

Verify the maximum value

For 6 workbooks:

For 7 workbooks:

Therefore, 6 is the greatest permissible number.

01.7.15 Common Mistakes in Mixed Problems

Mixed problems are challenging because students may correctly solve the individual equations or inequalities but combine their solutions incorrectly.

Mistake 1: Confusing union and intersection

Suppose:

If the question asks for \(A\cap B\), the answer is:

If it asks for \(A\cup B\), the answer is:

These are very different solution sets.

Mistake 2: Ignoring endpoint inclusion

Suppose:

There is no common value because 5 is excluded from \(B\).

Therefore:

Even though the intervals touch at 5 on a number line, their intersection is empty.

Mistake 3: Forgetting a rational denominator restriction

Consider:

The expression simplifies to \(x+2\), but only when \(x\neq2\).

Therefore, the answer is:

not \([-2,\infty)\).

Mistake 4: Selecting quadratic roots instead of intervals

Solve:

The roots are 2 and 4.

But the solution set is:

The roots identify boundaries; they are not automatically the solution set.

Mistake 5: Treating an integer problem as a real-number problem

Suppose:

If \(x\in\mathbb R\), the solution is:

If \(x\in\mathbb Z\), the solution is:

The number domain changes the form of the final answer.

Mistake 6: Forgetting that a system can have no solution

Consider:

and:

No real number satisfies both conditions.

Therefore:

Students should not automatically join these two intervals using a union symbol.

Final answer checklist

Before submitting any mixed problem, students should verify:

• Have all individual inequalities been solved correctly?

• Does the question require a union or an intersection?

• Are all interval endpoints included or excluded correctly?

• Have all denominator zeros been excluded?

• Have integer or other domain restrictions been applied?

• Does the final answer satisfy every original condition?

01.7.17 Lesson Summary

This mixed practice lesson connects the concepts learned throughout Module 01.

The most important ideas are:

1. Sets allow us to describe and combine collections of possible solutions.

2. Union includes elements belonging to at least one set.

3. Intersection includes only elements common to all required sets.

4. Linear inequalities are solved using equivalent algebraic operations.

5. Quadratic inequalities require locating roots and determining the sign of an expression.

6. Rational inequalities require sign analysis and strict attention to denominator restrictions.

7. Interval notation expresses solution sets compactly and precisely.

8. Additional restrictions can reduce a solution set or make it empty.

9. Integer restrictions require selecting only the permitted integer values.

10. Correct mathematical reasoning is more important than choosing an answer based only on the roots.

Essential formulas

Inclusion-exclusion for two finite sets:

Intersection of solution sets:

Union of solution sets:

Rational expression restriction:

General solution strategy:

Solve each mathematical condition, represent its solution as a set, combine the sets according to the logical requirements, and verify the final domain and endpoints.

This completes the written review portion of Lesson 01.7.