Simultaneous Equations
Solving linear simultaneous equations is a foundational skill assumed for Queensland Year 11 Mathematical Methods (QCAA). These are two linear equations that must both hold at once, and their solution is the single point where the two lines cross.
You will learn to solve a pair by substitution and by elimination, recognise when parallel lines give no solution, and connect the algebra to the point of intersection on a graph.
Theory
In Year 11 Mathematical Methods (QCAA), simultaneous equations are two linear equations that must both be true at once. This page reviews how to solve a pair by substitution and by elimination, and how the solution is the single point of intersection where the two lines cross. Parallel lines never meet, so such a pair has no solution.
A pair of simultaneous equations is two linear equations in the same two unknowns, for example \(x+y=7\) and \(x-y=3\). The solution is the pair \((x,\,y)\) that satisfies both equations at the same time.
Substitution makes one variable the subject of one equation and substitutes it into the other, leaving a single equation in one unknown. Elimination adds or subtracts multiples of the equations to remove one variable, again leaving one unknown.
Graphically, each equation is a straight line, and the solution is their point of intersection. If the lines are parallel (equal gradients, different intercepts) they never cross, so there is no solution.
For a pair in the same two unknowns, elimination scales the equations so one variable’s coefficients match, then adds or subtracts:
Substitution uses one equation solved for a variable, say
How to solve a pair of simultaneous equations
- Choose a method: substitution suits a pair where one variable is already alone; elimination suits equations lined up in the same form.
- Reduce to one equation in one unknown — substitute, or scale and add/subtract to eliminate a variable.
- Solve that equation, then back-substitute to find the other variable.
- Check the pair \((x,\,y)\) in both original equations.
Add the equations — the \(y\) terms cancel:
| \((x+y)+(x-y)\) | \(=\) | \(7+3\) |
| \(2x\) | \(=\) | \(10\) |
| \(x\) | \(=\) | \(5\) |
Back-substitute \(x=5\) into \(x+y=7\):
| \(5+y\) | \(=\) | \(7\) |
| \(y\) | \(=\) | \(2\) |
Check in \(x-y=3\):
| \(5-2\) | \(=\) | \(3\;\checkmark\) |
Solution: \((x,\,y)=(5,\,2)\).
Substitute \(y=2x-1\) into \(3x+y=9\):
| \(3x+(2x-1)\) | \(=\) | \(9\) |
| \(5x-1\) | \(=\) | \(9\) |
Solve for \(x\):
| \(5x\) | \(=\) | \(10\) |
| \(x\) | \(=\) | \(2\) |
Back-substitute into \(y=2x-1\):
| \(y\) | \(=\) | \(2(2)-1\) |
| \(=\) | \(3\) |
Check in \(3x+y=9\):
| \(3(2)+3\) | \(=\) | \(9\;\checkmark\) |
Solution: \((x,\,y)=(2,\,3)\).
Match the \(x\) terms — multiply the first by \(2\), the second by \(3\):
| \(6x+8y\) | \(=\) | \(20\) |
| \(6x+9y\) | \(=\) | \(21\) |
Subtract the first new equation from the second:
| \((6x+9y)-(6x+8y)\) | \(=\) | \(21-20\) |
| \(y\) | \(=\) | \(1\) |
Back-substitute \(y=1\) into \(3x+4y=10\):
| \(3x+4(1)\) | \(=\) | \(10\) |
| \(3x\) | \(=\) | \(6\) |
| \(x\) | \(=\) | \(2\) |
Check in \(2x+3y=7\):
| \(2(2)+3(1)\) | \(=\) | \(7\;\checkmark\) |
Solution: \((x,\,y)=(2,\,1)\).
Eliminate \(y\) — multiply the first by \(2\):
| \(14x-4y\) | \(=\) | \(6\) |
Add this to \(3x+4y=11\):
| \((14x-4y)+(3x+4y)\) | \(=\) | \(6+11\) |
| \(17x\) | \(=\) | \(17\) |
| \(x\) | \(=\) | \(1\) |
Back-substitute \(x=1\) into \(3x+4y=11\):
| \(3(1)+4y\) | \(=\) | \(11\) |
| \(4y\) | \(=\) | \(8\) |
| \(y\) | \(=\) | \(2\) |
Form the required quantity:
| \(x+y\) | \(=\) | \(1+2\) |
| \(=\) | \(3\) |
\(x+y=3\).
Common pitfalls
Frequently asked questions
What are simultaneous equations?
Two linear equations in the same two unknowns that must both be true at once. The solution is the pair \((x,\,y)\) satisfying both.
When should I use substitution instead of elimination?
Use substitution when one variable is already the subject (like \(y=2x-1\)); use elimination when the equations are lined up in the form \(ax+by=c\).
How does elimination work?
Scale the equations so one variable has matching coefficients, then add or subtract to remove it, leaving one equation in one unknown.
What does the solution mean graphically?
Each equation is a straight line, and the solution is their point of intersection — the one point that lies on both lines.
What if there is no solution?
If the two lines are parallel (equal gradients, different intercepts) they never meet, so the pair has no solution. Eliminating a variable then leaves a false statement such as \(0=6\).
How do I check my answer?
Substitute the pair \((x,\,y)\) into both original equations and confirm each is satisfied.