Year 11 Methods (Unit 1 & 2) Coordinate Geometry And Linear Relations

Simultaneous Linear Equations

Q: What does the solution of a linear system represent graphically?

The point where the two lines intersect. Its coordinates satisfy both equations at once.

Q: When should I use substitution rather than elimination?

Use substitution when one variable is already the subject or easily isolated. Use elimination when both equations are in the form a x plus b y equals c.

Q: How do I know how many solutions a system has?

Compare the gradients. Different gradients give one solution, equal gradients with different intercepts give none because the lines are parallel, and identical lines give infinitely many.

Q: How can I check my answer?

Substitute the solution back into both original equations. It must make each one true. If either fails, re-check the working.

20 practice questions 1 video lesson Theory + worked examples

Solving linear simultaneous equations is a foundational skill assumed for Queensland Year 11 Mathematical Methods (QCAA). A pair is solved by the values that satisfy both equations at once — graphically, the point where the two lines meet.

You will solve a pair by substitution and by elimination, connect the algebra to the point of intersection, and use gradients to decide whether there is one solution, none, or infinitely many.

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Theory

In Year 11 Mathematical Methods (QCAA), a pair of simultaneous linear equations is solved by the values of \(x\) and \(y\) that satisfy both. Algebraically use substitution or elimination; graphically the solution is the point of intersection of the two lines. This page shows both methods and how gradients tell you whether there is one solution, none, or infinitely many.

A pair of simultaneous linear equations has a solution \((x,y)\) that makes both equations true at once. Substitution makes one variable the subject of one equation and replaces it in the other. Elimination adds or subtracts multiples of the equations to cancel a variable.

Graphically, each equation is a line, and the solution is the point of intersection. The gradients tell the story: different gradients give exactly one solution; equal gradients with different intercepts are parallel (no solution); identical lines are coincident (infinitely many solutions).

Substitute when a variable is already isolated; eliminate when the equations line up in \(ax+by=c\) form. Both give the same intersection point.

The solution of a system is where the two lines cross.

Equal gradients, different intercepts: parallel lines, no solution.

For a system in the form

\[a_1x+b_1y=c_1,\qquad a_2x+b_2y=c_2\]

a_{1} x + b_{1} y = c_{1}

the number of solutions depends on the gradients:

\[m_1\neq m_2\Rightarrow\text{one},\quad m_1=m_2,\ c_1\neq c_2\Rightarrow\text{none},\quad \text{same line}\Rightarrow\infty\]

m_{1} \neq m_{2}

Check your solution: substitute \((x,y)\) back into both original equations — it must satisfy each one.

How to solve a linear system

Choose a method: substitution if a variable is (or is easily made) the subject; elimination if the equations are in \(ax+by=c\) form.
Reduce to one variable: substitute, or add/subtract suitable multiples to cancel a variable, then solve.
Back-substitute to find the other variable, state the point \((x,y)\), and check it in both equations.

Example 1 — Solve by substitution

Solve \(y=4x\) and \(2x+y=12\), and state the point of intersection.

Solution

Substitute \(y=4x\) into the second equation:

\(2x+(4x)\)	\(=\)	\(12\)
\(6x\)	\(=\)	\(12\)
\(x\)	\(=\)	\(2\)

Back-substitute to find \(y\):

\(y\)	\(=\)	\(4(2)\)
	\(=\)	\(8\)

Solution \((2,\,8)\) — the lines meet at \((2,\,8)\).

(2, 8)

Example 2 — Solve by elimination

Solve \(2x+3y=19\) and \(3x+2y=21\).

Solution

Make the \(x\)-coefficients match: multiply the first by \(3\), the second by \(2\):

\(6x+9y\)	\(=\)	\(57\)
\(6x+4y\)	\(=\)	\(42\)

Subtract to eliminate \(x\):

\(5y\)	\(=\)	\(15\)
\(y\)	\(=\)	\(3\)

Back-substitute into \(2x+3y=19\):

\(2x+3(3)\)	\(=\)	\(19\)
\(2x+9\)	\(=\)	\(19\)
\(2x\)	\(=\)	\(10\)
\(x\)	\(=\)	\(5\)

Solution \((5,\,3)\).

(5, 3)

Example 3 — Number of solutions from gradients

How many solutions does the system \(y=2x+5\) and \(y=2x-3\) have?

Solution

Read the gradient and intercept of each line:

\(m_1\)	\(=\)	\(2\)
\(m_2\)	\(=\)	\(2\)

The gradients are equal but the intercepts differ (\(5\neq -3\)), so the lines are parallel.

Try to solve — set the right-hand sides equal:

\(2x+5\)	\(=\)	\(2x-3\)
\(5\)	\(=\)	\(-3\)

This is impossible, confirming there is no intersection.

No solution — the lines are parallel.

no solution

Example 4 — A derived quantity

Solve \(5x+2y=16\) and \(3x-2y=0\), then find \(x+y\).

Solution

Add the equations to eliminate \(y\) (the \(+2y\) and \(-2y\) cancel):

\(8x\)	\(=\)	\(16\)
\(x\)	\(=\)	\(2\)

Back-substitute into \(3x-2y=0\):

\(3(2)-2y\)	\(=\)	\(0\)
\(6-2y\)	\(=\)	\(0\)
\(2y\)	\(=\)	\(6\)
\(y\)	\(=\)	\(3\)

Compute the required quantity:

\(x+y\)	\(=\)	\(2+3\)
	\(=\)	\(5\)

Solution \((2,\,3)\), so \(x+y=5\).

x + y = 5

Common pitfalls

Only finding one variable. A solution is a point \((x,y)\); after solving for one variable you must back-substitute for the other.

Adding when you should subtract. Elimination cancels a variable only if its coefficients are equal and opposite. Match them first, then add (opposite signs) or subtract (same signs).

Assuming every system has one solution. Parallel lines give no solution; coincident lines give infinitely many. Compare gradients before concluding.

Frequently asked questions

What does the solution of a linear system represent graphically?

The point where the two lines intersect. Its coordinates \((x,y)\) satisfy both equations at once.

When should I use substitution rather than elimination?

Use substitution when one variable is already the subject (or easily isolated); use elimination when both equations are in \(ax+by=c\) form.

How do I know how many solutions a system has?

Compare gradients: different gradients give one solution, equal gradients with different intercepts give none (parallel), and identical lines give infinitely many.

Why do parallel lines give no solution?

They have the same gradient but never meet, so there is no point \((x,y)\) that lies on both lines.

How can I check my answer?

Substitute the solution back into both original equations; it must make each one true. If either fails, re-check the working.

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