Solving Linear Inequalities for the ESAT

Symbols and Labelling Conventions

Inequalities use specific symbols to define ranges of values. When representing these on a number line, the style of circle used indicates whether the boundary value is included in the solution set.

1. Less than ($<$): $x < 4$ defines points such that $x$ can take any value less than 4, but not including 4. On a number line, this is shown with an open circle at 4.
2. Greater than ($>$): $x > 4$ defines points such that $x$ can take any value greater than 4, but not including 4. This is also shown with an open circle at 4.
3. Less than or equal to ($\leq$): $x \leq 4$ includes every value less than 4 and the value 4 itself. This is shown with a solid circle at 4.
4. Greater than or equal to ($\geq$): $x \geq 4$ includes every value greater than 4 and the value 4 itself. This is shown with a solid circle at 4.

Inequalities in Two Variables

When working with two variables, inequalities are represented as shaded regions on a coordinate graph. The boundary is defined by the equivalent linear equation. A dotted line indicates that the boundary is not included (for $<$ or $>$), while a solid line indicates the boundary is included (for $\leq$ or $\geq$).

In the graph above, the shaded region represents $x < y$. Because it is a strict inequality, the boundary line $x = y$ is dotted.

In this second graph, the shaded region shows $x \geq y$. The boundary line $x = y$ is solid because the values on the line satisfy the inequality.

Rules for Simplifying Inequalities

To solve or simplify an inequality, follow these algebraic rules to ensure the inequality sign remains correct:

• Addition and Subtraction: Adding or subtracting any real number from both sides leaves the sign unchanged. If $x < y$, then $x - 4 < y - 4$.
• Positive Multiplication and Division: Multiplying or dividing by a positive real number leaves the sign unchanged. If $x < y$, then $\frac{3}{4}x < \frac{3}{4}y$.
• Inversion: If $x$ and $y$ are both positive or both negative, inverting both sides changes the direction of the sign. If $x < y$, then $\frac{1}{x} > \frac{1}{y}$.
• Negative Multiplication and Division: Multiplying or dividing by a negative real number changes the direction of the inequality sign. If $x < y$, then $-\frac{3}{4}x > -\frac{3}{4}y$.

Combining Inequalities

Combined inequalities take the form $a < b < c$ or $a \leq b \leq c$. Both signs must point in the same direction. For instance, $a < b > c$ is mathematically incorrect as a combined statement.

• $x > -7$ and $x \leq 10$ can be written as $-7 < x \leq 10$.
• $x < -7$ and $x \leq 10$ simplify to the single range $x < -7$.
• $x < -7$ and $x \geq 10$ do not define a single continuous range and cannot be combined into one statement.
• The statement $-7 < x < -10$ is invalid because it implies $-7 < -10$, which is false.

Worked Examples: One Variable

Simple Linear Inequality

Solve $3x - 4 > 8$ and show the result on a number line.

1. Add 4 to both sides: $3x > 12$
2. Divide by 3: $x > 4$

The number line would feature an unfilled circle at 4 with an arrow pointing to the right.

Two Sided Linear Inequality

Solve $20 \geq 3x - 4 > 8$ and show the result on a number line.

1. Add 4 to all three sections: $24 \geq 3x > 12$
2. Divide all three sections by 3: $8 \geq x > 4$

This is represented by an unfilled circle at 4 and a filled circle at 8, with a line connecting them.

Finding a Range from Two Inequalities

Find the range for which both $3x - 4 < 8$ and $3x - 2 > -8$ are valid.

1. Solve the first: $3x < 12 \implies x < 4$
2. Solve the second: $3x > -6 \implies x > -2$
3. Combine the results: $-2 < x < 4$

Complicated Inequalities

Solve $3x - 4 < 2(2 + 3x)$. (Note: Following the guide's step-by-step logic for the similar expression $3x - 5 < 2(2 + 3x)$):

1. Multiply out the bracket: $3x - 5 < 4 + 6x$
2. Subtract $6x$ from both sides: $-3x - 5 < 4$
3. Add 5 to both sides: $-3x < 9$
4. Multiply by -1 and reverse the sign: $3x > -9$
5. Divide by 3: $x > -3$

Worked Examples: Two Variables

Graphing a Single Inequality

Shade the region for $3x + 4y \leq 12$.

1. Draw the boundary line $3x + 4y = 12$ as a solid line.
2. Test a point. At $(0, 0)$, $3(0) + 4(0) = 0$, which is less than 12. Therefore, the region containing the origin is the correct one.

Simultaneous Inequalities Graphically

Draw the region where $3x + 4y \leq 12$, $y \geq x + 6$, and $x > -7$ are all satisfied.

To find the solution, it is often easier to shade out the unwanted regions:

1. For $3x + 4y \leq 12$, shade the region above the solid line.
2. For $y \geq x + 6$, shade the region below the solid line.
3. For $x > -7$, shade the region to the left of the broken line.

The unshaded area remaining is the required solution.