Solving Inequalities Revision Notes for HSC SSCE Mathematics Advanced

Solving Inequalities

What are inequations?

When we replace the equals sign in an equation with an inequality symbol like $<$ , $\leq$ , $>$ , or $\geq$ , we create an inequation.

For example, the equation $3x = 12$ has the solution $x = 4$ . If we change this to the inequation $3x > 12$ , the solution becomes $x > 4$ . This is because any number greater than $4$ makes the statement true, whilst any other number makes it false.

Inequations vs inequalities

There is an important distinction to understand:

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Understanding the Terminology

An inequation is a statement like $3x > 12$ that is only true for certain values of $x$
An inequality is a statement like $x^2 \geq 0$ that is true for all values of $x$

Think of inequalities as similar to identities in equations. For instance, $(x + 3)^2 = x^2 + 6x + 9$ is true for all values of $x$ .

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Different textbooks may use these terms differently. Sometimes "inequality" is used for both concepts, so don't be confused if you're asked to "solve an inequality" - it means the same as solving an inequation.

Understanding "less than"

The phrase "less than" can be understood in two helpful ways. For real numbers $a$ and $b$ :

Geometric interpretation

We say that $a < b$ if $a$ is to the left of $b$ on the number line.

Algebraic interpretation

We say that $a < b$ if $b - a$ is positive.

Both interpretations are useful tools when solving inequations. The geometric interpretation helps us visualise solutions, whilst the algebraic interpretation is helpful for algebraic manipulation.

Solving linear inequations

The methods for solving linear inequations are very similar to those for solving linear equations, with one crucial difference.

Key rules

Add or subtract the same value from both sides (just like equations)
Multiply or divide both sides by the same value (just like equations)
Never multiply or divide by $0$

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Critical Rule: Reversing the Inequality

When multiplying or dividing both sides by a negative number, you must reverse the inequality symbol.

For example: If $-2x < 6$ , then dividing by $-2$ gives $x > -3$ (note the sign reversal).

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Worked Example: Solving Linear Inequations

Example (a): Solve $3x - 7 \leq 8x + 18$

Solution process:

$3x - 7 \leq 8x + 18$

Add $(-8x + 7)$ to both sides:

$-5x \leq 25$

Divide by $(-5)$ and reverse the inequality:

$x \geq -5$

Example (b): Solve $20 > 2 - 3x \geq 8$

This is a compound inequality. We need to work with both parts simultaneously.

$20 > 2 - 3x \geq 8$

Subtract $2$ from all parts:

$18 > -3x \geq 6$

Divide by $(-3)$ and reverse both inequalities:

$-6 < x \leq -2$

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Exam Tip

Always remember to flip the inequality sign when dividing or multiplying by a negative number. This is one of the most common mistakes in examinations.

Solving quadratic inequations

The most effective way to solve a quadratic inequation is to sketch the graph of the associated parabola. This graphical method makes the solution clear and visual.

Method for quadratic inequations

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Step-by-Step Approach

Move everything to the left-hand side
Sketch the graph of $y = \text{LHS}$ , showing the x-intercepts
Read the solution off the graph

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Worked Examples: Quadratic Inequations

Example (a): Solve $x^2 > 9$

Step 1: Move everything to the left:

$x^2 - 9 > 0$

Step 2: Factorise:

$(x - 3)(x + 3) > 0$

Step 3: Sketch $y = (x - 3)(x + 3)$ and identify where the graph is above the x-axis.

The parabola is above the x-axis when $x > 3$ or $x < -3$ .

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This example is straightforward and can be solved by recognising that $x^2 > 9$ means $x$ is further than $3$ units from zero.

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Example (b): Solve $x + 6 \geq x^2$

Step 1: Move everything to the left:

$x^2 - x - 6 \leq 0$

Step 2: Factorise:

$(x - 3)(x + 2) \leq 0$

Step 3: Sketch $y = (x - 3)(x + 2)$ and identify where the graph is below or on the x-axis.

The parabola is below or on the x-axis when $-2 \leq x \leq 3$ .

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Exam Tip

For quadratic inequations, always sketch the parabola. The visual representation helps prevent sign errors and makes the solution region clear.

Solving absolute value equations and inequations

Absolute value problems can be solved using two main approaches: the number line method (for simple cases) and the algebraic method (for more complex cases).

Using the number line

Most absolute value equations and inequations in this course can be solved by thinking about distances on the number line.

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Key Principle

$|x - a|$ represents the distance from $x$ to $a$ on the number line.

Method for simple absolute value problems

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Step-by-Step Approach

Rewrite the equation or inequation in the form $|x - b| = a$ , or $|x - b| < a$ , or $|x - b| \geq a$
Interpret this as a distance on the number line
Find the solution by considering which values of $x$ satisfy the distance condition

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Worked Examples: Number Line Method

Example (a): Solve $|x - 2| = 5$

This asks: "What values of $x$ are distance $5$ from $2$ ?"

On the number line, we see that $x = -3$ or $x = 7$ .

Example (b): Solve $|x + 3| = 4$

Rewrite as: "What values of $x$ are distance $4$ from $-3$ ?"

This gives $x = -7$ or $x = 1$ .

Example (c): Solve $|3x + 7| < 3$

First, divide by $3$ :

$\left|x + 2\frac{1}{3}\right| < 1$

This asks: "What values of $x$ are less than distance $1$ from $-2\frac{1}{3}$ ?"

This gives $-3\frac{1}{3} < x < -1\frac{1}{3}$ .

Example (d): Solve $|7 - \frac{1}{4}x| \geq 3$

Multiply by $4$ :

$|28 - x| \geq 12$

This asks: "What values of $x$ are at least distance $12$ from $28$ ?"

This gives $x \leq 16$ or $x \geq 40$ .

Algebraic method

For more complex absolute value problems, we can use algebraic rewriting.

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Algebraic Rewriting Rules

Absolute value equations:

Rewrite $|f(x)| = a$ as: $f(x) = a \text{ or } f(x) = -a$

Absolute value inequations:

Rewrite $|f(x)| < a$ as: $-a < f(x) < a$

Rewrite $|f(x)| > a$ as: $f(x) < -a \text{ or } f(x) > a$

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Special Case: Negative Values of $a$

The absolute value $|f(x)|$ cannot be negative. Therefore, if $a$ is negative:

$|f(x)| = a$ and $|f(x)| < a$ have no solutions
$|f(x)| > a$ is true for all values of $x$ in the domain of $f(x)$

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Worked Example: Algebraic Method

Consider $|9 - 2x|$ with different conditions:

Part (a): Solve $|9 - 2x| = 5$

Rewrite as two equations:

$9 - 2x = 5 \text{ or } 9 - 2x = -5$

Subtract $9$ :

$-2x = -4 \text{ or } -2x = -14$

Divide by $(-2)$ :

$x = 2 \text{ or } x = 7$

Part (b): Solve $|9 - 2x| < 5$

Rewrite as a compound inequality:

$-5 < 9 - 2x < 5$

Subtract $9$ :

$-14 < -2x < -4$

Divide by $(-2)$ and reverse inequalities:

$7 > x > 2$

This can be written as: $2 < x < 7$

Part (c): Solve $|9 - 2x| > 5$

Rewrite as two separate inequalities:

$9 - 2x < -5 \text{ or } 9 - 2x > 5$

Subtract $9$ :

$-2x < -14 \text{ or } -2x > -4$

Divide by $(-2)$ and reverse inequalities:

$x > 7 \text{ or } x < 2$

This can be written as: $x < 2$ or $x > 7$

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Exam Tip

When solving absolute value inequations, pay careful attention to whether you need "and" (for $<$ problems) or "or" (for $>$ problems) in your final answer.

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Key Points to Remember

Reversing the inequality: When multiplying or dividing both sides of an inequation by a negative number, always reverse the inequality symbol
Quadratic inequations: Sketch the parabola and identify where it's above or below the x-axis to find your solution
Absolute value as distance: Think of $|x - a|$ as the distance from $x$ to $a$ on the number line
Algebraic rewriting: For $|f(x)| < a$ , rewrite as $-a < f(x) < a$ . For $|f(x)| > a$ , rewrite as $f(x) < -a$ or $f(x) > a$
Check your work: Test a value from your solution in the original inequation to verify your answer is correct

Solving Inequalities (HSC SSCE Mathematics Advanced): Revision Notes