Solve systems of simultaneous linear equations in two and three variables, including by substitution, elimination, and matrix methods, and interpret the results graphically

What this dot point is asking

QCAA wants you to solve systems of simultaneous linear equations in two and three variables, by substitution, elimination or matrix methods, and to interpret the geometric meaning of the solutions.

Why a system has three possible outcomes

A single linear equation in two unknowns has infinitely many solutions (every point on its line), so a unique answer needs a second, independent equation. Geometrically each equation is a line, and solving the system means finding their common points. Because two lines in a plane either cross once, run parallel, or coincide, exactly three outcomes are possible. The algebra mirrors this: a consistent independent pair gives one solution, an inconsistent pair (parallel lines) gives none, and a dependent pair (the same line written twice) gives infinitely many.

Two-variable systems

A pair of linear equations represents two lines in the plane. Three cases:

• Unique solution. Lines intersect at exactly one point.
• No solution. Lines are parallel.
• Infinitely many solutions. Lines coincide.

Substitution method

1. Solve one equation for one variable.
2. Substitute into the other equation.
3. Solve the resulting single-variable equation.
4. Back-substitute to find the other variable.

Elimination method

1. Multiply equations to align coefficients of one variable.
2. Add or subtract to eliminate that variable.
3. Solve for the remaining variable.
4. Back-substitute.

Graphical interpretation

The solution point is the intersection of the lines. If lines have the same gradient but different intercepts: parallel, no solution. If equations are scalar multiples of each other: identical lines, infinite solutions.

Three-variable systems

Three equations in three unknowns represent three planes. Possible outcomes:

• Unique point of intersection.
• Line of intersection (infinitely many solutions).
• No common point.

Use elimination repeatedly to reduce to a $2 \times 2$ system, then solve.

Matrix methods (Year 11 introduction)

For $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} e \\ f \end{pmatrix}$, the solution is $\begin{pmatrix} x \\ y \end{pmatrix} = A^{-1} \begin{pmatrix} e \\ f \end{pmatrix}$ where $A^{-1} = \dfrac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.

If $\det A = ad - bc = 0$, the matrix has no inverse: lines are parallel or coincident.

Choosing a method

Substitution is cleanest when one equation already has a variable with coefficient $1$ (easy to isolate); elimination is cleanest when coefficients line up or can be matched by a single multiplication. Matrix inversion is efficient when the same coefficient matrix is reused with different right-hand sides. All three give the same answer, so pick the one that minimises fractions for the numbers in front of you.

In one sentence

Simultaneous linear equations are solved by substitution (isolate one variable, substitute, solve) or elimination (align coefficients, add or subtract), with three possible outcomes (unique intersection, parallel lines with no solution, coincident lines with infinite solutions) interpreted graphically and algebraically through the determinant in matrix form.