Additional Mathematics
06062025–2027 syllabus

ADDITIONAL MATHEMATICS · CHAPTER 5

Simultaneous equations

Reduce two linked conditions to one equation, then recover and verify every ordered pair.

Additional Mathematics4 connected sectionsSyllabus-aligned guide

LEARNING OBJECTIVES

What you will be able to do

  • choose between elimination and substitution
  • solve linear–nonlinear systems
  • handle product and reciprocal relationships
  • check solutions and interpret intersections

AT A GLANCE

Syllabus0606Coverage2025–2027Sections4LevelAdditional Mathematics

INTRODUCTION · THE BIG IDEA

Reduce two linked conditions to one equation, then recover and verify every ordered pair.

Simultaneous equations ask for values that satisfy both conditions at once. In Additional Mathematics, one or both equations may be nonlinear, so substitution often creates a quadratic or higher equation.

Each real solution for one variable may generate a matching value for the other. The answer is a set of ordered pairs, not two unrelated lists.

01

SECTION 01

Choosing an elimination route

Core concept

Use elimination when matching terms can be added or subtracted. Use substitution when one variable, product or reciprocal expression can be isolated cleanly.

Method choice
StructureEfficient start
two linear equationseliminate a variable
one variable already isolatedsubstitute directly
xy is knownreplace the repeated product
reciprocal expressionclear denominators after restrictions
line with a curvesubstitute the line into the curve
02

SECTION 02

Linear and quadratic systems

Core concept

Substituting a line into a quadratic relation gives one quadratic equation. Its discriminant also predicts whether the graphs meet twice, touch once or do not meet in real coordinates.

RULE 1
two intersections: Δ > 0
RULE 2
tangent contact: Δ = 0
RULE 3
no real intersection: Δ < 0
Original worked example

Intersect a line and a parabola

  1. Solve y = x + 1 with x² + y² = 13.
  2. Substitute: x² + (x + 1)² = 13.
  3. Simplify: x² + x − 6 = 0, so x = 2 or x = −3.
  4. Use y = x + 1 to obtain y = 3 or y = −2.

Answer: (2, 3) and (−3, −2).

03

SECTION 03

Products, powers and reciprocals

Core concept

If a relation such as xy = k is available, use y = k/x with x ≠ 0 or replace xy wherever it appears. Equations containing y/x and x/y should be multiplied by a valid common denominator.

DETAILED EXPLANATION

  • If xy² and xy are known, division may reveal y.
  • A substitution can produce powers such as y²; recover both signs where valid.
  • Check that no multiplication by zero introduced a candidate.
ORIGINAL STUDY DIAGRAMSolve a nonlinear pair
1Record restrictions
2Isolate a useful expression
3Substitute into the second equation
4Solve the resulting equation
5Recover and verify each pair
04

SECTION 04

Checking and presenting solutions

Core concept

Substitution, squaring and clearing denominators can introduce extraneous results. Test each pair in the original equations, then present paired values together.

DETAILED EXPLANATION

  • Complex solutions are not required unless a question explicitly changes the number system.
  • Graphical solutions should be read to the requested accuracy.
  • Reject a pair if either original equation fails.

QUICK CHAPTER SUMMARY

The ideas to carry forward

  • A simultaneous solution satisfies all equations.
  • Elimination suits matching terms; substitution suits an isolated expression.
  • Nonlinear systems can have several, one or no real pairs.
  • Restrictions must be stated before clearing denominators.
  • Every candidate pair should be checked in the original system.

QUICK REVISION CHECKLIST

Can you do each of these without your notes?

  • choose between elimination and substitution
  • solve linear–nonlinear systems
  • handle product and reciprocal relationships
  • check solutions and interpret intersections