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Simultaneous Equations

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Cambridge (CIE) IGCSE · Additional Maths

Simultaneous Equations

The big idea: when you've got two unknowns hiding in two clues, you combine the clues — by eliminating one letter or substituting one equation into the other — until only one unknown is left standing.

Summary — everything in this chapter, at a glance
  • Linear simultaneous equations only have x and y to the power 1 (no , , or xy). Solve them with the elimination method or the substitution method.
  • Quadratic (non-linear) simultaneous equations have an , , or xy in at least one equation. These must be solved by substitution — you always substitute the linear equation into the quadratic one, never the other way round.
  • Graphically, solving simultaneous equations = finding where two graphs cross. The coordinates of the crossing point(s) are the solution(s).
  • Quadratic simultaneous equations can have 0, 1, or 2 solutions — because a line can miss a curve, touch it (tangent), or cut through it twice.
  • Worded problems are just simultaneous equations wearing a costume — the hard part is translating the words into two equations correctly.
  • Always check your final answers in both original equations. This catches almost every sign error.
1. What Are Simultaneous Equations?

Two unknowns need two clues

Imagine someone tells you: "I'm thinking of two numbers. Their sum is 11." You genuinely can't figure out what those numbers are — there are infinite pairs that sum to 11 (1 and 10, 2 and 9, 5.5 and 5.5...). One clue for two unknowns just isn't enough information.

But now they tell you a second clue: "...and their difference is 3." Suddenly you can pin down exactly one pair of numbers. That's the whole idea behind simultaneous equations: when you have two unknowns, you need two equations (two independent pieces of information) to find unique values for both.

"Simultaneous" literally means "at the same time" — you're looking for the pair of values (an x and a y) that make both equations true simultaneously, not just one of them.

Example pair 3x + 2y = 11   and   2x − y = 5
Solution: x = 3, y = 1 — and this is the only pair that satisfies both equations at once.

Linear vs. Quadratic — how to tell them apart instantly

Look at the powers of x and y in each equation:

TypeWhat you'll seeExample
LinearOnly x and y — no squares, no xy3x + 2y = 11
Quadratic / Non-linearAn , , or an xy term somewherex² + y² = 25

If you graph a linear equation, you get a straight line (linear = line — easy way to remember it). If you graph a quadratic equation like x² + y² = 25 you get a curve (in that case, a circle). This matters because two straight lines cross at most once, but a line and a curve can cross 0, 1, or 2 times — which is exactly why quadratic simultaneous equations can give you two different answer-pairs.

2. Linear Equations — The Elimination Method
Make one unknown disappear, solve for the other, then backtrack.

The core idea: cancel one letter out

Elimination works like this: if two equations both contain, say, 6x, and you subtract one equation from the other, the x's vanish completely (6x − 6x = 0). You're left with a single equation in a single unknown — which you can just solve directly.

The elimination recipe 1. Multiply one or both equations by a constant so that the coefficients of x (or y) match.
2. If the matching terms have the same signsubtract. If they have different signsadd.
3. Solve the resulting one-unknown equation.
4. Substitute that value back into either original equation to find the other unknown.
5. Check both original equations.
Plain English: line the two equations up so one letter matches in size, then combine them so that letter cancels.

Worked example — start to finish

Solve: −2x + 4y = 5   and   4x − 5y = −7
① −2x + 4y = 5
② 4x − 5y = −7
Step 1 — match a coefficient. Multiply ① by 2: −4x + 8y = 10
Now compare the x-terms: −4x (from the new ①) and +4x (from ②) — different signs, so we add.
Step 2 — add. (−4x + 8y) + (4x − 5y) = 10 + (−7)  →  3y = 3
Step 3 — solve. y = 1
Step 4 — substitute back into equation ②: 4x − 5(1) = −7  →  4x = −2  →  x = −0.5
Step 5 — check in equation ①: −2(−0.5) + 4(1) = 1 + 4 = 5 ✓
Answer: x = −0.5, y = 1
Why the same-sign / different-sign rule works Think of it like a tug-of-war. If both equations pull the matched term the same way (both +6x, say), subtracting cancels them because you're removing the exact same amount twice. If they pull in opposite directions (+6x and −6x), adding cancels them because they directly cancel each other out, like +5 and −5 summing to 0.
Practice Question 1
Solve using elimination:   5x + 3y = 19   and   2x − 3y = 9
Practice Question 2
Solve using elimination:   4x + 3y = 2   and   3x + 2y = 3
3. Linear Equations — The Substitution Method
Rearrange one equation, then plug it into the other.

Swap a letter for an expression

Substitution takes a different route to the same destination. Instead of cancelling a term, you rearrange one equation so that y (or x) is completely on its own — "the subject" of the equation. Then, wherever y appears in the other equation, you replace it with that expression. This turns two equations with two unknowns into one equation with just one unknown.

The substitution recipe 1. Rearrange one equation into the form y = ... (or x = ...).
2. Substitute that expression into the other equation, wrapping it in brackets.
3. Solve the resulting one-unknown equation.
4. Substitute that value back into your rearranged equation to find the other unknown.
5. Check both original equations.

Worked example

Solve: 3x + 2y = 11   and   2x − y = 5
Step 1. Rearrange the second equation: y = 2x − 5
Step 2. Substitute into the first equation: 3x + 2(2x − 5) = 11
Step 3. Expand and solve: 3x + 4x − 10 = 11 → 7x = 21 → x = 3
Step 4. Substitute back: y = 2(3) − 5 = 1
Step 5. Check: 3(3) + 2(1) = 11 ✓ and 2(3) − 1 = 5 ✓
Answer: x = 3, y = 1
Elimination vs. substitution — which do I use? For linear pairs, either method always works — pick whichever is less messy. Elimination tends to be faster when coefficients are easy to match. Substitution is quicker when one equation already has an unknown on its own (like y = 1 − 9x). But here's the crucial bit: for quadratic simultaneous equations, you have no choice — substitution is the only method that works, so it's worth being fully comfortable with it now.
Practice Question
Solve using substitution:   3x + 5y = −9   and   y = 1 − 9x
4. Solving Graphically

The solution is where lines cross

Every equation in x and y can be drawn as a graph. A linear equation draws a straight line; every point on that line makes the equation true. If you draw two equations on the same axes, then any point where the two graphs intersect is a point that satisfies both equations at once — which is exactly what "solving simultaneously" means.

        y
        |        2x−y=3
        |          /
        |         /
        |    ★(2,1)  ← lines cross here
        |       /  \
        |      /    \
        |_____/______\____ x
        |    /        3x+y=7
      

Reading off the graph: the lines above cross at (2, 1), so the solution to 2x − y = 3 and 3x + y = 7 simultaneously is x = 2, y = 1.

Why this matters for quadratics This graphical view is what makes it click why quadratic simultaneous equations can have two answers: picture a straight line cutting through a parabola. It can miss the curve entirely (0 solutions), just graze it at one point (1 solution — a tangent), or slice straight through it (2 solutions). Algebra just finds those crossing points without needing to actually draw anything.
5. Solving Worded Problems
The maths is easy once you've translated the words correctly.

Turning a story into two equations

Worded questions are simultaneous equations in disguise. The skill being tested isn't really the algebra — it's spotting that there are two unknown quantities and two independent pieces of information, and translating each piece of information into an equation. Always start by clearly defining your letters, in words, before writing any equations.

Worked example

At a fireworks stall, a customer pays £9 for six bangers and twelve sparklers. Another customer buys nine bangers and ten sparklers for £12.30. Find the cost of 5 bangers and 15 sparklers.
Define letters: let B = price of a banger (£), S = price of a sparkler (£)
Translate each sentence into an equation:
① 6B + 12S = 9
② 9B + 10S = 12.3
Eliminate B: ①×3: 18B + 36S = 27     ②×2: 18B + 20S = 24.6
Subtract: 16S = 2.4 → S = 0.15
Substitute into ①×3: 18B + 36(0.15) = 27 → 18B = 21.6 → B = 1.2
Answer the actual question asked (not just B and S!):
5B + 15S = 5(1.2) + 15(0.15) = 6 + 2.25 = 8.25
Five bangers and fifteen sparklers cost £8.25
Don't stop too early! Notice the question never actually asked for the price of one banger or one sparkler — it asked for the cost of a completely different combination (5 bangers + 15 sparklers). Always re-read the question after solving for your unknowns to check whether there's one final calculation left to do.
Practice Question
A cinema sells adult tickets for £A and child tickets for £C. One family buys 2 adult and 3 child tickets for £29. Another family buys 3 adult and 1 child ticket for £26. Find the cost of 1 adult ticket and 1 child ticket.
6. Quadratic Simultaneous Equations
One linear, one quadratic — substitution is the only way in.

Why elimination won't work here

Elimination relies on lining up matching terms and cancelling them by adding or subtracting. That's only possible when both equations have the same "shape" of terms. Once one equation has an or xy and the other doesn't, there's nothing to match up — so elimination is off the table. Substitution, however, always works because you're not trying to match anything; you're replacing a whole variable with an expression, however complicated that expression is.

The golden rule Always substitute the linear equation into the quadratic one. Never try to substitute the quadratic equation into the linear one — it creates a much messier and often unsolvable equation.

Worked example — full method

Solve: x² + y² = 36   and   x = 2y + 6
① x² + y² = 36 (quadratic)     ② x = 2y + 6 (linear, already rearranged for x)
Step 1 — substitute ② into ①: (2y + 6)² + y² = 36
Step 2 — expand carefully. Treat (2y+6)² exactly like (2y+6)(2y+6):
4y² + 12y + 12y + 36 + y² = 36
Step 3 — simplify: 5y² + 24y + 36 = 36
Step 4 — rearrange to equal zero: 5y² + 24y = 0
Step 5 — factorise (common factor of y): y(5y + 24) = 0
Step 6 — solve for both roots: y₁ = 0    or    5y₂ + 24 = 0 → y₂ = −24/5 = −4.8
Step 7 — substitute each y back into the linear equation (it's easier):
x₁ = 2(0) + 6 = 6     x₂ = 2(−4.8) + 6 = −9.6 + 6 = −3.6
Answer: x = 6, y = 0   OR   x = −3.6, y = −4.8
Always pair your answers correctly Notice there are two complete solution pairs here, not four separate loose numbers. Mixing up which x goes with which y is one of the most common ways to lose marks — write your final answer as clearly matched pairs, like (x₁, y₁) = (6, 0) and (x₂, y₂) = (−3.6, −4.8).

Reading the number of solutions from the discriminant

Once you've substituted and simplified down to a quadratic equation (something you'd solve with factorising, completing the square, or the quadratic formula), the discriminant tells you geometrically what's happening between the line and the curve:

DiscriminantNumber of solutionsGeometric meaning
Positive (b² − 4ac > 0)2 solutionsThe line cuts through the curve at two points
Zero (b² − 4ac = 0)1 repeated solutionThe line is a tangent — it just touches the curve
Negative (b² − 4ac < 0)0 real solutionsThe line completely misses the curve
Practice Question
Solve the simultaneous equations:   y = x² − 2x − 1   and   y = x + 5
What to Memorise
Term / RuleWhat it means
Simultaneous equationsTwo (or more) equations that must be true at the same time — solved together to find unique values for each unknown.
Linear equationOnly powers of 1 for x and y (no x², y², xy). Graphs as a straight line.
Quadratic / non-linearContains x², y², or xy. Graphs as a curve.
Elimination methodMatch coefficients, then add (different signs) or subtract (same signs) to cancel one variable.
Substitution methodRearrange one equation for x or y, then substitute that expression into the other equation.
Golden rule for quadraticsAlways substitute the linear equation into the quadratic one — never the reverse.
Graphical solutionSolutions = the coordinates of the point(s) where the two graphs intersect.
Tangent caseA repeated root (discriminant = 0) means the line touches the curve at exactly one point.
No solution caseA negative discriminant means the line and curve never meet.
Checking stepSubstitute your final x and y values into both original equations to confirm they work.
Concepts Checklist
Exam Tips
Don't skip the checking step It only takes a few seconds, and there are so many places to make a small error when solving simultaneous equations. Substituting your final answers back into both original equations is the single fastest way to catch mistakes before you hand in your paper.
Watch your minus signs Mishandling minus signs is probably the single biggest cause of student error in this topic — especially when deciding whether to add or subtract during elimination, or when expanding brackets with negative terms. Slow down on this step specifically.
Label your equations Number your equations ① and ② right at the start. It sounds trivial, but it makes it far easier to reference "substitute ② into ①" without confusing yourself midway through a multi-step solution — examiners also reward clear, trackable working.
Substitute into the easier equation When finding your second unknown, plug your value back into whichever original equation is simpler to rearrange — usually the linear one if you're working on a quadratic pair. This reduces arithmetic errors.
State which x goes with which y For quadratic simultaneous equations, if you don't clearly show which x-value pairs with which y-value, you can lose marks even if both numbers are individually correct. Write your final answer as matched pairs, e.g. "x = 6, y = 0 or x = −3.6, y = −4.8."
Re-read worded questions before finishing In worded problems, the final answer often requires one more calculation after you've found your two unknowns (like the "5 bangers and 15 sparklers" example). Always re-check exactly what quantity the question is actually asking for.
Cambridge (CIE) IGCSE Additional Maths  ·  Simultaneous Equations Revision Guide
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