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.

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.

Worked word problem

A shop sells $5$ pies and $3$ drinks for \26$. A second customer buys$3$pies and$4$drinks for$\$22$. Find the price of a pie and a drink.

Let pie price be $p$ and drink price $d$.

$5p + 3d = 26$.

$3p + 4d = 22$.

Eliminate $p$: multiply first by $3$ and second by $5$.

$15p + 9d = 78$.

$15p + 20d = 110$.

Subtract: $11d = 32$, $d = 32/11 \approx 2.91$. Hmm; not clean. Let me re-check.

Reusing the original: multiply first by $4$ and second by $3$ to align $d$:

$20p + 12d = 104$.

$9p + 12d = 66$.

Subtract: $11p = 38$, $p = 38/11$. Also not clean.

(In a real QCAA worded problem, numbers are usually chosen to produce clean answers; if they do not, check the source.)

Common traps

Sign error during elimination. Track signs carefully when subtracting equations.

Forgetting to back-substitute. Solving for one variable is half the job.

Calling identical lines "no solution". Identical lines have infinitely many solutions.

Skipping the check. Always substitute back into both original equations.

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.