1 · What Are Simultaneous Equations?
Imagine I tell you: "I'm thinking of two numbers. Their sum is 19." Can you tell me exactly
what the two numbers are? No — there are infinite possibilities (10 and 9, 15 and 4, 1 and 18...).
One equation with two unknowns has endless solutions.
But now I add a second clue: "...and their difference is 5." Suddenly, there's only
one pair of numbers that satisfies both clues at once: 12 and 7. That's
the whole idea of simultaneous equations — two unknowns need two equations to pin
down a single, unique answer (in the linear case).
Think of it like this
Each equation is a "rule" that a huge family of (x, y) pairs could satisfy. Solving simultaneously means finding the one pair (or pairs) that obeys every rule at once.
There are two types you'll meet in this chapter:
- Linear simultaneous equations — no squared terms or
xy terms. Example: 3x + 2y = 11 and 2x − y = 5.
- Quadratic (non-linear) simultaneous equations — at least one equation contains
x², y², or xy. Example: x² + y² = 25 and y − 2x = 5.
2 · Solving Linear Equations by Elimination
Elimination means making one of the letters disappear by combining the two equations
— either by adding them together or subtracting one from the other.
The Method, Step by Step
- Number your equations ① and ② — this keeps your working organised and stops you losing track.
- Match the coefficients of x (or y) in both equations. Multiply each entire equation by whatever number is needed. Whatever you do to one term, you must do to every term in that equation.
- Add or subtract the equations to eliminate that letter (use the golden rule above).
- Solve the resulting one-letter equation.
- Substitute that value back into either original equation to find the other letter.
- Check both values in the other original equation — this is your safety net against arithmetic slips.
Worked Example
Solve: 5x + 2y = 11 ① and 4x − 3y = 18 ②
1. We'll eliminate y. Multiply ① by 3 and ② by 2 so both y-coefficients become 6:
15x + 6y = 33 ③ 8x − 6y = 36 ④
2. The 6y terms have opposite signs (+6y and −6y) → ADD ③ and ④:
23x = 69 → x = 3
3. Substitute x = 3 into ①: 5(3) + 2y = 11 → 15 + 2y = 11 → y = −2
4. Check in ②: 4(3) − 3(−2) = 12 + 6 = 18 ✓ Correct!
Solution: x = 3, y = −2
Practice Question 1
Solve the simultaneous equations: 3x + 2y = 11 and 2x − y = 5
Practice Question 2
Solve: 6b + 12s = 9 and 9b + 10s = 12.3
3 · Solving Linear Equations by Substitution
Substitution is an alternative to elimination — some students find it more intuitive.
Instead of cancelling a letter out, you replace it entirely.
The Method, Step by Step
- Rearrange one equation so it reads
y = ... (or x = ...).
- Substitute that expression into the other (unaltered) equation, replacing every occurrence of that letter — always in brackets, to avoid sign errors.
- Solve the resulting one-letter equation.
- Substitute that value back into your rearranged equation from Step 1 to find the second letter.
Worked Example
Solve: 3x + 2y = 11 and 2x − y = 5
1. Rearrange the second equation: y = 2x − 5
2. Substitute into the first: 3x + 2(2x − 5) = 11
3. Expand and solve: 3x + 4x − 10 = 11 → 7x = 21 → x = 3
4. Substitute x = 3 into y = 2x − 5: y = 6 − 5 = 1
Solution: x = 3, y = 1 — same answer as elimination, different route!
Elimination vs Substitution — which should I use?
- Elimination is usually faster and cleaner when both equations are already in the form
ax + by = c.
- Substitution is essential when one equation is already solved for a letter, or — as you'll see in Topic 5 — when one equation is quadratic.
Practice Question
Use substitution to solve: y = x + 2 and 2x + 3y = 16
4 · Solving Graphically
Every straight line is a picture of every (x, y) pair that satisfies its equation. So if you draw
two lines on the same axes, the point where they cross is the one
(x, y) pair that satisfies both equations — exactly what "solving simultaneously" means!
For example, to solve 2x − y = 3 and 3x + y = 4 simultaneously: plot both
lines, and you'll find they cross at (2, 1). So the solution is x = 2, y = 1
— no algebra needed, just careful plotting and reading.
Watch out
The graphical method is only as accurate as your drawing. If the intersection point doesn't land on a "nice" grid coordinate, you may only be able to estimate the solution — algebra (elimination/substitution) gives you the exact answer.
5 · Forming Simultaneous Equations from Words
Exam questions often hide the equations inside a real-world story. Your job is to translate the words
into algebra before you even start solving.
The Method
- Define your letters clearly — e.g. "let b = the price of a bagel in pence." Always state units.
- Build two separate equations from the two pieces of information given.
- Solve simultaneously using elimination or substitution.
- Answer in context — go back to the original story and state what your numbers actually mean (with correct units). Don't just stop at "x = 40, y = 30"!
Worked Example
A customer pays £9 for six bagels and twelve sausage rolls. Another pays £12.30 for nine bagels and ten sausage rolls. Find the cost of 5 bagels and 15 sausage rolls.
1. Let b = price of a bagel (£), s = price of a sausage roll (£).
2. Equations: 6b + 12s = 9 ① and 9b + 10s = 12.3 ②
3. Solve simultaneously (see Practice Question 2 above) → b = 1.2, s = 0.15
4. Answer in context: bagels cost £1.20, sausage rolls cost £0.15. So 5 bagels + 15 sausage rolls = (5 × 1.2) + (15 × 0.15) = £8.25
Not every worded question needs "simultaneous" solving explicitly
Example: "Two numbers have a sum of 19 and a difference of 5 — find their product." You still form x + y = 19 and x − y = 5, solve to get x = 12, y = 7, then go one step further: xy = 12 × 7 = 84.
Practice Question
3 apples and 2 bananas cost £1.80. 5 apples and 1 banana cost £2.30. Find the price of one apple and one banana.
6 · Quadratic (Non-Linear) Simultaneous Equations
Everything changes slightly once one of your equations has an x², y², or
xy term. Picture it visually: a linear equation is a straight line, but
a quadratic equation is a curve (like a parabola or a circle). A straight line can
cross a curve at two points, one point (just grazing it — called a
tangent), or not at all.
The Method, Step by Step
- Number your equations. Identify which one is linear and which is quadratic.
- Rearrange the linear equation into
y = ... or x = ... (pick whichever avoids creating a fraction — this keeps the algebra cleaner).
- Substitute this expression into the quadratic equation, replacing every occurrence of that letter (in brackets!).
- Expand and simplify — you'll now have a normal quadratic equation in one letter.
- Solve the quadratic (factorise, or use the quadratic formula) — this typically gives two values.
- Substitute each value back into the rearranged linear equation to find its matching partner value.
- Pair up your answers correctly — this is the #1 place marks are lost!
Worked Example
Solve: x² + y² = 36 ① and x − 2y = 6 ②
1. Rearrange ② to make x the subject (avoids a fraction): x = 2y + 6
2. Substitute into ①: (2y + 6)² + y² = 36
3. Expand (2y+6)² like double brackets: 4y² + 24y + 36 + y² = 36
4. Simplify: 5y² + 24y = 0 (the 36's cancel)
5. Factorise: y(5y + 24) = 0 → y = 0 or y = −24/5
6. Substitute both into x = 2y + 6:
y = 0 → x = 6 | y = −24/5 → x = −18/5
Solutions: (x = 6, y = 0) and (x = −18/5, y = −24/5)
What if the equation isn't already "y = ..." or "x = ..."?
You'll need to rearrange it first. For example, to solve xy = 3 and x + y = 4, rearrange the linear one to y = 4 − x, then substitute into xy = 3: x(4 − x) = 3, which expands into the quadratic 4x − x² = 3.
How Many Solutions Will I Get?
Once you substitute and simplify, you'll land on a quadratic equation. Look at its roots to know what's happening geometrically:
- Two distinct roots → the line crosses the curve at two points → two solution pairs.
- One repeated root → the line just touches the curve at one point (it's a tangent) → one solution pair.
- No real roots → the line never meets the curve → no solutions. (If this happens unexpectedly, double-check your algebra — it's often a sign of an error!)
Practice Question
Solve the simultaneous equations: y = x² + 3x + 1 and y = 2x + 1
Practice Question
Solve: x² + y² = 25 and y − 2x = 5