Simultaneous Equations
Two equations, two unknowns, and one pair of values that satisfies both. These are typically 4- or 5-mark questions where the marks are spread across the method, so working is worth writing out even if the arithmetic goes wrong.
The errors here are almost entirely sign errors during subtraction — not conceptual confusion.
1. The elimination method
Solve: 3x + 2y = 16 and 5x − 2y = 8
The y terms already have equal coefficients with opposite signs, so adding removes them:
- (3x + 2y) + (5x − 2y) = 16 + 8
- 8x = 24 → x = 3
- Substitute into the first equation: 3(3) + 2y = 16 → 2y = 7 → y = 3.5
- x = 3, y = 3.5
Same signs subtract, different signs add
If the matching coefficients have the SAME sign, SUBTRACT. If they have DIFFERENT signs, ADD.
A memory hook used in lessons: Same Signs Subtract.
Example needing subtraction: 4x + 3y = 18 and 2x + 3y = 12
- The +3y terms match with the same sign → subtract
- (4x + 3y) − (2x + 3y) = 18 − 12
- 2x = 6 → x = 3, then 4(3) + 3y = 18 → y = 2
Use brackets when subtracting, and subtract EVERY term. This is where most marks in the topic are lost. Writing (4x + 3y) − (2x + 3y) makes it visible that 3y − 3y = 0; without brackets students frequently end up with 3y, or lose the sign on the constant.
A term subtracted from itself gives 0, not itself. “3y − 3y = 3y” and “3y − 3y = y” were both recorded errors.
Making the coefficients match
Usually neither pair matches, so multiply one or both equations first.
Solve: 2x + 3y = 12 and 5x − 2y = 11
To eliminate y, make the coefficients 6:
- Equation 1 × 2: 4x + 6y = 24
- Equation 2 × 3: 15x − 6y = 33
- Signs differ → add: 19x = 57 → x = 3
- Substitute: 2(3) + 3y = 12 → 3y = 6 → y = 2
Multiply EVERY term, including the number on the right. Multiplying only the left-hand side is a frequent slip.
Choose the variable that is easiest to match — look for coefficients where one divides into the other, or where the signs already differ.
2. The substitution method
Better when one equation already has a variable on its own.
Solve: y = 2x + 1 and 3x + y = 11
- Substitute the first into the second: 3x + (2x + 1) = 11
- 5x + 1 = 11 → 5x = 10 → x = 2
- y = 2(2) + 1 = 5
Use brackets when you substitute. It protects the signs, especially when the expression starts with a minus.
3. Both methods: the finishing steps
After finding the first value, SUBSTITUTE to find the second — don’t run elimination a second time. Trying to eliminate again was a documented error and wastes time.
Substitute into the simpler original equation, and then check in the other one. That check costs seconds and catches nearly every arithmetic error.
Pair your answers clearly: x = 3, y = 2. Giving two loose numbers without saying which is which loses the final mark. If the question used other letters, or you defined your own, state what they mean — “12 adult tickets and 7 child tickets”.
4. Solving graphically
The solution is where the two lines intersect.
- Rearrange both to y = mx + c
- Draw both lines
- Read off the coordinates of the crossing point
The intersection satisfies both equations at once — that is exactly what “simultaneous” means. If a question says “use the graph”, you must read the values off; solving algebraically instead can score nothing.
Special cases:
- Parallel lines never meet → no solution
- Identical lines overlap → infinitely many solutions
5. Quadratic simultaneous equations (Extended)
When one equation is linear and the other quadratic, always use substitution.
Solve: y = x + 1 and x² + y² = 25
- Rearrange the LINEAR equation and substitute into the quadratic: x² + (x + 1)² = 25
- Expand carefully: x² + x² + 2x + 1 = 25
- 2x² + 2x − 24 = 0 → divide by 2: x² + x − 12 = 0
- Factorise: (x + 4)(x − 3) = 0 → x = −4 or x = 3
- Substitute each back into y = x + 1:
- x = −4 → y = −3
- x = 3 → y = 4
- (−4, −3) and (3, 4)
Substitute the linear equation into the quadratic, not the other way round. Rearranging the quadratic instead was a recorded error and makes the algebra much worse.
(x + 1)² is x² + 2x + 1, not x² + 1. This single expansion error accounts for a large share of lost marks here.
There are usually TWO solution pairs, and you must give both — each x with its own y. Pair them correctly; mixing them up gives points that don’t lie on either curve.
Substitute back into the LINEAR equation to find y. It is far quicker and less error-prone than using the quadratic.
If it doesn’t factorise, use the quadratic formula.
6. Word problems
Example: 3 adult and 2 child tickets cost 32. 1 adult and 4 child tickets cost 24. Find each price.
- Define the variables: let an adult ticket be a, a child ticket c
- Form equations: 3a + 2c = 32 and a + 4c = 24
- Solve: multiply the second by 3 → 3a + 12c = 72; subtract → 10c = 40 → c = 4; then a + 16 = 24 → a = 8
- Adult tickets cost 8, child tickets cost 4
Define your variables in words before you start, and answer in words at the end. Tutors flagged both; the final statement is often worth a mark.
If the question says “use simultaneous equations”, you must use them — even if you can see the answer another way. A correct answer by trial and error can score zero when a method is specified.
7. Mistakes that cost marks
Adding when you should subtract, or vice versa.
Not using brackets when subtracting equations.
Failing to subtract every term, especially the constant.
Getting 3y − 3y wrong.
Multiplying only one side of an equation.
Running elimination twice instead of substituting.
Not checking in the second equation.
Giving unpaired values, or not saying which is which.
Expanding (x + 1)² as x² + 1.
Substituting the quadratic into the linear.
Giving only one solution pair for a quadratic system.
Mismatching x and y across the two pairs.
Solving algebraically when the question said “use the graph”.
Frequently asked questions
What are simultaneous equations? Two equations with two unknowns, solved together to find the pair of values satisfying both.
When do I add and when do I subtract? Same signs subtract, different signs add — for the matching coefficients.
How do I make the coefficients match? Multiply one or both equations by whatever is needed — multiplying every term.
When should I use substitution? When one equation already gives a variable on its own, and always for quadratic simultaneous equations.
How do I find the second variable? Substitute the value you found into one of the original equations.
How do I check my answer? Put both values into the other equation and confirm it balances.
What does the solution mean graphically? The point where the two lines intersect.
What if the lines are parallel? No solution.
How many answers do quadratic simultaneous equations have? Usually two pairs — give both, correctly paired.
Which equation do I substitute back into? The linear one — it is much simpler.
Quick revision checklist
- I can solve by elimination, adding or subtracting correctly
- I remember same signs subtract
- I use brackets when subtracting equations
- I multiply every term when scaling an equation
- I substitute to find the second variable rather than eliminating twice
- I check my values in the other equation
- I present answers as a clearly labelled pair
- I can solve by substitution, using brackets
- I can find the solution graphically and know what intersection means
- I know parallel lines mean no solution
- I substitute the linear equation into the quadratic
- I expand squared brackets correctly
- I give both solution pairs, correctly matched
- I define variables in words and answer in words
- I use the method the question demands
These notes cover simultaneous equations in the Cambridge IGCSE Mathematics (0580) syllabus, and are written for Grade 9–11 / Year 10–11 students. They are based on teaching patterns observed across a large set of one-to-one IGCSE Maths lessons, with particular attention to the errors students make most often and the wording examiners reward. Always check the current syllabus and formula list for your own exam series.
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