Solving Simultaneous Equations
Solve two equations together using substitution or elimination. Start with two linear equations, then try the clearly labelled Higher example with a quadratic.
What are simultaneous equations?
Simultaneous equations are when two equations are solved together to find the values that work in both equations at the same time.
We start with substitution because it shows how one equation helps solve the other. Then we use elimination to remove an unknown by adding or subtracting equations.
Example Question:
We will work through this pair step by step:
Solve the simultaneous equations
Exam Tip
Always rearrange the equation that is easiest to work with first.
Look for:
- A variable with coefficient 1
- An equation that is already nearly rearranged
This reduces the chance of mistakes.
Method – Solving Simultaneous Equations by Substitution
The substitution method solves simultaneous equations by replacing one variable with an expression from the other equation.
Follow these steps:
- Rearrange one equation
Choose the equation that is easiest to rearrange and make one variable the subject. You are isolating one variable so it can be substituted into the other equation. - Substitute into the second equation
Replace that variable in the second equation with the expression you found. This creates a new equation with only one unknown. - Solve the new equation
Simplify, expand brackets if needed, and solve to find the value of the remaining variable. - Substitute back
Substitute this value into one of the original equations to find the other variable. - Check your solution
Substitute both values into the original equations to confirm they work. This step is strongly recommended in exams.
Worked Example
Solve:
Step 1: Rearrange the first equation:
We’ll rearrange the simpler equation to make the subject.
(Subtract from both sides)
Step 2: Substitute into the second equation:
Now substitute into the second equation:
(Replace with .)
Step 3: Expand and simplify
Expand the bracket:
Combine like terms:
(
Step 4: Solve for
Subtract 10 from both sides:
Multiply both sides by
(Be careful with negative signs here.)
Step 5: Substitute back to find
Use the rearranged equation:
Substitute :
Final answer:
The solution to the simultaneous equations is:
Always give both values – not just one.
Check your solution (important habit)
Substitute the values into both original equations.
First equation:
✓
Second equation:
✓
Both equations are satisfied, so the solution is correct.
Choosing a method and extending your skills
Worked example — Eliminate matching terms
Solve 2x + y = 11 and x + y = 7. The y terms are equal, so subtract the entire second equation from the first.
(2x + y) − (x + y) = 11 − 7.
x = 4.
Substitute into x + y = 7: 4 + y = 7, so y = 3.
Check both: 2 × 4 + 3 = 11 and 4 + 3 = 7.
If the terms have opposite signs, add instead: x + y = 7 and x − y = 1 give 2x = 8.
Worked example — Make coefficients match
Solve 2x + 3y = 19 and 3x + 2y = 16.
Multiply every term of the first equation by 3: 6x + 9y = 57.
Multiply every term of the second by 2: 6x + 4y = 32.
Subtract: 5y = 25, so y = 5.
Substitute into 2x + 3y = 19: 2x + 15 = 19.
2x = 4, so x = 2, y = 5.
Check the second equation: 3 × 2 + 2 × 5 = 16.
Worked example — Form the equations
Two adult tickets and one child ticket cost £19. One adult ticket and two child tickets cost £17. Find each price.
Let a be an adult ticket price in pounds and c a child ticket price. Then 2a + c = 19 and a + 2c = 17.
Double the second equation: 2a + 4c = 34.
Subtract the first equation: 3c = 15, so c = 5.
Use 2a + 5 = 19: 2a = 14, so a = 7.
Adult £7; child £5. Checks: 2 × 7 + 5 = 19 and 7 + 2 × 5 = 17.
Graphical meaning — Foundation and Higher
Each linear equation describes a straight line. A point on both lines satisfies both equations, so their crossing gives the solution.
For x + y = 7 and x − y = 1, write y = 7 − x and y = x − 1.
For the first line, plot (0, 7), (4, 3) and (7, 0). For the second, plot (0, −1), (1, 0) and (4, 3).
Draw both lines on the same axes. They cross at (4, 3), so x = 4 and y = 3.
A solution read from a graph is approximate unless its coordinates are known exactly. Check it in both equations.
Parallel distinct lines have no common solution. If two equations describe the same line, there are infinitely many common solutions.
Higher extension — One linear and one quadratic equation
Solve y = x + 2 and y = x². This example uses quadratic factorising; learn that skill first.
Both expressions equal y, so x² = x + 2.
Subtract x and 2: x² − x − 2 = 0.
Factorise: (x − 2)(x + 1) = 0.
A product is zero when at least one factor is zero. Thus x = 2 or x = −1.
For x = 2, y = 2 + 2 = 4. For x = −1, y = −1 + 2 = 1.
The solution pairs are (2, 4) and (−1, 1).
Check in y = x²: 2² = 4 and (−1)² = 1. Keep each x with its matching y.
Higher — a linear equation and a circle equation
When the quadratic contains both x² and y², substitute for one variable and solve the resulting quadratic. The circle equation is supplied here; circle-equation and tangent teaching belongs to its later lesson.
Worked example — Solve y = x + 1 and x² + y² = 25
Substitute the entire expression x + 1 for y: x² + (x + 1)² = 25.
Expand: 2x² + 2x + 1 = 25. Subtract 25 and divide by 2: x² + x − 12 = 0.
Factorise: (x + 4)(x − 3) = 0. Thus x = −4 or x = 3. Using y = x + 1 gives (−4, −3) and (3, 4).
Check both original equations for both pairs: −3 = −4 + 1 and 16 + 9 = 25; 4 = 3 + 1 and 9 + 16 = 25. Do not mix coordinates from different pairs.
Higher practice — Solve y = x − 1 and x² + y² = 13. Check both pairs.
Substitute: x² + (x − 1)² = 13. Hence 2x² − 2x − 12 = 0, so x² − x − 6 = (x − 3)(x + 2) = 0.
x = 3 or −2. The paired solutions are (3, 2) and (−2, −3).
Checks: 2 = 3 − 1 and 9 + 4 = 13; −3 = −2 − 1 and 4 + 9 = 13.
Think Like a Mathematician
Rearranging equations can feel unfamiliar at first, but it follows clear algebra rules.
You are simply isolating one variable and reducing the problem step by step until only one unknown remains.
The order of the equations does not matter — logical steps do.
Common mistakes to avoid
- Forgetting to substitute into the entire equation
- Missing brackets when substituting
- Errors when expanding brackets
- Finding one variable but forgetting to substitute back
- Sign errors when simplifying
Most mistakes occur during substitution and expansion — slow down at these steps.
Try one yourself
Solve the following pair:
Hint: Rearrange either equation first – both are equally easy this time.
From x + y = 7, subtract y to get x = 7 − y.
Substitute this into x − y = 1: (7 − y) − y = 1.
Collect terms: 7 − 2y = 1. Subtract 7 from both sides: −2y = −6.
Divide both sides by −2: y = 3.
Substitute y = 3 into x = 7 − y: x = 4.
Check both equations: 4 + 3 = 7 and 4 − 3 = 1.
Answer: x = 4, y = 3.
More practice — choose and check a method
1. Solve x + y = 9 and x − y = 3.
Add the equations: 2x = 12, so x = 6.
Use 6 + y = 9: y = 3.
x = 6, y = 3. Check: 6 + 3 = 9 and 6 − 3 = 3.
2. Solve 2x + y = 12 and x − y = 3.
Add the equations: 3x = 15, so x = 5.
Use 5 − y = 3: y = 2.
x = 5, y = 2. Check: 2 × 5 + 2 = 12 and 5 − 2 = 3.
3. Solve 3x + 2y = 16 and 2x + 3y = 19.
Multiply the first by 2: 6x + 4y = 32.
Multiply the second by 3: 6x + 9y = 57.
Subtract: 5y = 25, so y = 5.
3x + 10 = 16 gives x = 2. Checks: 6 + 10 = 16 and 4 + 15 = 19.
4. Two pens and a notebook cost £8. A pen and a notebook cost £5. Find both prices.
Let p be the pen price and n the notebook price, in pounds.
2p + n = 8 and p + n = 5. Subtract to get p = 3.
Then n = 5 − 3 = 2. Pen £3; notebook £2.
Check: 2 × 3 + 2 = 8 and 3 + 2 = 5.
5. The lines y = 2x + 1 and y = 7 − x cross at (2, 5). Explain what this tells you and check it.
The same point is on both lines, so the simultaneous solution is x = 2, y = 5.
Check: 2 × 2 + 1 = 5 and 7 − 2 = 5.
6. Higher: solve y = x² and y = 2x + 3.
Equate the expressions: x² = 2x + 3.
x² − 2x − 3 = 0, so (x − 3)(x + 1) = 0.
x = 3 or x = −1.
Using y = 2x + 3 gives (x, y) = (3, 9) or (−1, 1).
Check: 3² = 9 and (−1)² = 1.
Frequently Asked Questions
What is the substitution method?
The substitution method involves rearranging one equation and substituting it into the other to form a single equation with one variable.
When should I use substitution instead of elimination?
Substitution is usually easier when one equation already has a variable with coefficient 1 or is easy to rearrange.