Linear Inequalities and Simultaneous Equations — AC9M10A02

What you will learn

• solve linear inequalities and represent their solution sets
• reverse an inequality correctly when multiplying or dividing by a negative
• graph inequalities in 2 variables using a boundary line and test point
• solve simultaneous linear equations by elimination and substitution
• interpret the intersection of two lines as a simultaneous solution
• translate algebraic solutions back into the real situation

Prerequisite knowledge

You should be able to solve linear equations, rearrange simple formulas, substitute ordered pairs, plot straight-line graphs and interpret gradient and intercept. For simultaneous equations, accurate sign handling and collecting like terms are essential.

Concept teaching

1. An inequality describes a set

The statement x < 4 means every real number less than 4 is a solution. A strict sign (< or >) excludes the boundary; ≤ or ≥ includes it. When multiplying or dividing both sides by a negative number, reverse the inequality because the order of numbers on the number line reverses under multiplication by a negative.

2. Inequalities in 2 variables describe regions

For 2x + 3y < 12, first draw the boundary 2x + 3y = 12. Because the inequality is strict, the boundary is not included. Test a point not on the line, often (0,0). Since 0 < 12 is true, the solution region is the side containing the origin.

3. A simultaneous solution satisfies both equations

If two straight-line equations are both true, their common ordered pair is where the lines intersect. Algebraically, elimination combines equations to remove one variable; substitution replaces one variable with an equivalent expression. Graphing makes the meaning visible but may give only an approximate solution unless the intersection is clear.

4. Method choice matters

Use substitution when a variable is already isolated or easy to isolate. Use elimination when coefficients already match or can be made to match simply. Use graphs when interpretation, feasibility or an approximate intersection is central to the question.

Worked examples

Common misconceptions and corrections

• Forgetting to reverse the sign: only reverse it when multiplying or dividing both sides by a negative.
• Using a solid line for x+y<5: strict inequalities exclude the boundary, so use a dashed boundary.
• Shading by eye: use a test point unless the correct side is immediately justified.
• Finding x but not y: a simultaneous solution needs both coordinates.
• Eliminating the wrong way: if coefficients have the same sign, subtract; if opposite, add—or scale first.
• Ignoring context: negative or fractional solutions can be algebraically valid but impossible for counts or physical constraints.

Guided practice

1. Solve 5x−7>18.
2. Solve −4x≤20 and explain the sign change.
3. Solve x+y=12 and x−y=2 by elimination.
4. Solve y=3x−4 and x+y=12 by substitution.
5. For x+2y<8, state the boundary type and which side is shaded after testing (0,0).

Independent practice

1. Solve 7−2x≥15.
2. Solve −3<2x+5≤11.
3. Test whether (3,2) satisfies 2x+5y≤16.
4. Solve 2x+y=11 and x−y=1.
5. Solve 3x+2y=16 and x−2y=0.
6. Solve y=4x−5 and 2x+y=13.
7. A taxi plan costs 8+2.4d and another costs 14+1.8d. Find the distance d where costs are equal and interpret it.
8. Graphically describe the solution region x≥0, y≥0, x+2y≤10.

Reasoning and problem-solving task

A school club sells adult tickets for $18 and student tickets for $12. It can seat at most 240 people and wants at least $3600 revenue.

1. Let a and s be the numbers of adult and student tickets. Write two inequalities.
2. Explain what the first-quadrant restriction means.
3. Decide whether (120,100) is feasible.
4. Find two different feasible integer combinations.
5. Explain why this problem has a region of solutions rather than one simultaneous-equation solution.

Important questions and answers

What does an intersection mean?

It is an ordered pair satisfying both equations at the same time.

When should I use elimination?

When one variable has equal or easily matched coefficients, making it efficient to remove.

Why does an inequality sign reverse after dividing by a negative?

Multiplying by a negative reverses number order: for example 2<5 but −2>−5.

How do I choose which inequality region to shade?

Test a point not on the boundary and shade the side where the inequality is true.

Can simultaneous equations have no solution?

Yes. Distinct parallel lines never intersect. Coincident lines represent infinitely many common solutions.

Assessment-style questions

1. 3 marks: Solve −5(2x−1)<25 and represent the solution in inequality notation.
2. 4 marks: Solve 3x+2y=19 and 5x−2y=13 by elimination.
3. 4 marks: Solve y=2x−7 and 3x+y=13 by substitution and verify the ordered pair.
4. 5 marks: Draw or describe the solution region for x≥0, y≥0 and 2x+3y≤18. Explain the role of each boundary.
5. 6 marks: Two phone plans are C=35+0.08m and C=20+0.13m, where m is monthly minutes. Find the break-even point, determine which plan is cheaper below and above it, and state one practical limitation of this simple model.

Review hint: In an applied simultaneous-equations question, a correct ordered pair can still lose an interpretation mark. Finish with a sentence using the quantities and units from the problem.

Exit ticket: mastery check

1. Solve −3x>12.
2. What does a dashed boundary mean?
3. Solve x+y=8 and x−y=4.
4. State one situation where substitution is preferable to elimination.
5. Explain the meaning of a graph intersection in one sentence.

Teacher and parent guidance

For teachers

Connect methods rather than teaching them as unrelated recipes. After algebraic solving, graph the same pair so students see why the solution is an intersection. For inequalities, insist on verbal reasoning about boundaries, test points and context. Include no-solution and infinitely-many-solution cases once the main method is secure.

For parents and carers

If a student gets stuck, ask what each equation or inequality is saying before asking them to calculate. After they solve it, ask, “Does your answer satisfy both conditions?” This encourages checking rather than answer hunting.

Curriculum alignment

Australian Curriculum: AC9M10A02, Year 10 Algebra — solving linear inequalities and simultaneous linear equations in 2 variables, interpreting graphical solutions and communicating results in context.

Victoria: VC2M10A08, Level 10 Algebra covers linear inequalities, including number-line and Cartesian-plane representations; VC2M10A09 covers simultaneous linear equations using algebraic and graphical methods. Together they provide a strong alignment.

NSW: Stage 5 distributes this content mainly through Paths. MA5-EQU-P-01 includes linear inequalities, MA5-EQU-P-02 includes linear simultaneous equations, and MA5-FNC-P-01 supports graphing inequalities in one and 2 variables. MAO-WM-01 supports communicating and justifying solutions. This is not a single-outcome equivalence.

Alignment explanation: The lesson's one- and two-variable inequality work maps directly to the Australian descriptor and Victorian A08, while elimination/substitution/graph interpretation maps to Australian A02 and Victorian A09. NSW places much of the equivalent complexity in Stage 5 Paths, so the page identifies those Path outcomes explicitly rather than calling them Core.

Practice and teaching resources

Related Year 10 Mathematics topics

• AC9M10A01 — algebraic expressions and equations
• AC9M10A03 — exponential relations
• AC9M10A04 — mathematical modelling
• All Year 10 Mathematics topics

Official curriculum references

• ACARA — Australian Curriculum v9 Mathematics
• VCAA — Mathematics Version 2.0, Level 10
• NSW Curriculum/NESA — Mathematics K–10 (2022), Stage 5