International Mathematics (0607) | Inequalities

🌟 Algebra Chapter 2: Inequalities 🌟

Welcome to the chapter on Inequalities! You’ve spent a lot of time solving equations (where one side equals the other). But in the real world, things often aren't exactly equal. We deal with limits, budgets, and minimum requirements all the time.

This chapter teaches you how to mathematically describe situations where one quantity is greater than, less than, or at most another quantity. Mastery of inequalities is essential for problem-solving in mathematics and beyond!

Section 1: The Basics of Inequalities and Notation

An inequality is a mathematical statement that shows the relationship between two expressions that are not equal.

Key Inequality Symbols

Representing Inequalities on a Number Line (C2.6 / E2.6)

Since an inequality usually has many possible solutions (a range of numbers), we use a number line to show the solution set clearly. The notation used for the circles is very important:

1. Strict Inequalities (< or >): Use an Open Circle (or hollow circle).
   This shows that the number itself is not included in the solution set.

   Example: \(x > 2\). The number 2 is not a solution, but 2.000001 is.

2. Inclusive Inequalities (≤ or ≥): Use a Closed Circle (or solid circle).
   This shows that the number itself is included in the solution set.

   Example: \(x \leq 5\). The number 5 is a solution.

💡 Memory Aid:
The symbol \(\leq\) (less than or equal to) has a line underneath, like a solid base. Use a Solid (Closed) circle.
The symbol \(<\) (less than) is just open, so use an Open circle.

Example Representation:

\(x > -1\)
<-----O------|------|------|------|-----> x
      -2     -1      0     1     2

\(x \leq 3\)
<-----|------|------|------●-----> x
      1     2     3     4     5

Section 2: Solving Linear Inequalities (C2.6 / E2.6)

Solving a linear inequality is almost exactly the same as solving a linear equation. You use the same operations (add, subtract, multiply, divide) to isolate the variable (\(x\)).

The Golden Rule of Inequalities

There is one crucial difference that students often forget. It is the most common mistake made in this topic!

If you multiply or divide the entire inequality by a negative number, you must REVERSE the inequality sign.

Why? Let's use a simple analogy:
We know that \(5 > 2\). This is true.
If we multiply both sides by \(-1\):
\(-5\) and \(-2\).
Now, \(-5\) is actually smaller than \(-2\). Therefore, we must flip the sign: \(-5 < -2\).

Step-by-Step Solving Process

Example 1 (Basic Core Example): Solve \(5 - 2x \leq 13\)

1. Subtract 5 from both sides:
   \(5 - 2x - 5 \leq 13 - 5\)
   \(-2x \leq 8\)
2. Divide by -2 (Golden Rule Alert!):
   Since we are dividing by a negative number (\(-2\)), we must reverse the \(\leq\) sign to \(\geq\).
   \(\frac{-2x}{-2} \geq \frac{8}{-2}\)
   \(x \geq -4\)
3. Interpretation: The solution is all numbers greater than or equal to -4. (This would be represented by a closed circle at -4 and an arrow pointing right.)

Example 2 (Extended Construction Example): Solve \(3x < 2x + 4\)

1. Collect \(x\) terms on one side: Subtract \(2x\) from both sides.
   \(3x - 2x < 4\)
   \(x < 4\)
2. Interpretation: The solution is all numbers strictly less than 4. (Open circle at 4, arrow pointing left.)

Section 3: Compound Inequalities (Extended E2.6)

A compound inequality combines two simple inequalities into a single statement, usually representing a continuous range of values.

Example: \(-3 < x < 5\) means that \(x\) must be greater than \(-3\) AND less than \(5\).

Solving Compound Inequalities (The 'Sandwich' Method)

When solving inequalities like \(a \leq bx + c < d\), you perform the same operation on all three parts of the inequality simultaneously.

Example: Solve \(-3 \leq 3x - 2 < 7\)

1. Isolate the term with \(x\) in the middle. Start by adding 2 to all three parts:
   \(-3 + 2 \leq 3x - 2 + 2 < 7 + 2\)
   \(-1 \leq 3x < 9\)
2. Isolate \(x\). Divide all three parts by 3 (a positive number, so no sign flip needed):
   \(\frac{-1}{3} \leq \frac{3x}{3} < \frac{9}{3}\)
   \(-\frac{1}{3} \leq x < 3\)
3. Interpretation (Number Line): The solution is all values from \(-\frac{1}{3}\) (inclusive, closed circle) up to 3 (exclusive, open circle).

Did you know? Sometimes it's easier to split a compound inequality into two separate ones and solve them individually, then look for the overlap in their solutions. For example, \(-3 \leq 3x - 2 < 7\) can be solved as:
1. \(-3 \leq 3x - 2\)
2. \(3x - 2 < 7\)

Section 4: Graphical Inequalities (Extended E2.6)

When dealing with inequalities in two variables (like \(x\) and \(y\)), the solution is an entire region on the Cartesian plane, not just a line or a point.

Boundary Lines and Shading

First, you must draw the boundary line by replacing the inequality sign with an equals sign (e.g., \(y < 2x + 1\) becomes \(y = 2x + 1\)).

1. Strict Boundary (< or >): Use a Broken (Dotted) Line.
   (The points on the line are not solutions.)
2. Inclusive Boundary (≤ or ≥): Use a Solid Line.
   (The points on the line are solutions.)

Shading Convention (Crucial for IGCSE 0607):

Unless the question specifically directs otherwise, the standard convention in this syllabus (E2.6) is to shade the UNWANTED region. This means the clear, unshaded region represents the solution set.

Step-by-Step: Determining the Solution Region

Example: Find the region defined by \(y > 2x - 1\).

1. Draw the Boundary: Draw the line \(y = 2x - 1\). Because the inequality is \(>\) (strict), use a broken line.
2. Test a Point: Pick a simple test point, usually the origin \((0, 0)\), and substitute it into the original inequality:
   \(y > 2x - 1\)
   \(0 > 2(0) - 1\)
   \(0 > -1\)
3. Determine Shading:
   • Is \(0 > -1\) True? Yes. This means \((0, 0)\) is in the desired solution region.
   • Since we shade the unwanted region, we shade the side of the line that does NOT contain \((0, 0)\).

💪 Pro Tip for \(y > \text{or } y <\): If the inequality is written in the form \(y > mx + c\), the wanted region is above the line. If it’s \(y < mx + c\), the wanted region is below the line. Remember this rule *after* ensuring the coefficient of \(y\) is positive!

Defining a Region Given a Graph

Sometimes you are given a graph with a shaded region and asked to list the inequalities that define the unshaded (solution) region.

Process:

• Identify the equation of each boundary line (\(y = mx + c\) or \(x = k\)).
• Check if the line is solid (use \(\leq\) or \(\geq\)) or broken (use \(<\) or \(>\)).
• Use the origin \((0, 0)\) or another test point to determine the correct inequality direction relative to the unshaded region.

For example, if the boundary is \(y = 1\) (solid line) and the unshaded solution region is above it, the inequality is \(y \geq 1\).

Solving Equations/Inequalities Using a GDC (Extended E2.6.3)

Your Graphic Display Calculator (GDC) is a powerful tool for visualising and solving inequalities, especially those that might be unfamiliar or non-linear (though the syllabus focuses primarily on linear ones here).

To solve an inequality like \(2x^2 + 3 < 11\):

1. Graph \(y_1 = 2x^2 + 3\) and \(y_2 = 11\).
2. Find the intersection points of the two graphs. These points define the boundaries of the solution.
3. Look at the graph: where is the parabola \(y_1\) below the horizontal line \(y_2\)? The \(x\)-values in that segment are your solution.