What are simultaneous equations?
Simultaneous equations are two (or more) equations that are both true at the same time. The solution is the pair of values — usually an x value and a y value — that satisfies both equations simultaneously.
At GCSE you need three methods:
- Elimination — add or subtract the equations to remove one variable (the most common method at GCSE)
- Substitution — rearrange one equation and substitute into the other (essential for one linear, one quadratic)
- Graphical — draw both lines and read off the intersection point
Always check your final answer by substituting both values back into both original equations.
Elimination method
The goal is to make the coefficient of one variable the same in both equations, then add or subtract to eliminate it. If the signs are the same, subtract the equations. If the signs are opposite, add them.
Worked Example 1 — Simple elimination
Solve: 3x + 2y = 16 and 5x + 2y = 24
1
The coefficients of y are both +2 — subtract equation 1 from equation 2 to eliminate y: (5x + 2y) − (3x + 2y) = 24 − 16 → 2x = 8 → x = 4
2
Substitute x = 4 into equation 1: 3(4) + 2y = 16 → 2y = 4 → y = 2
3
Check in equation 2: 5(4) + 2(2) = 20 + 4 = 24 ✓
Answerx = 4, y = 2
Worked Example 2 — Elimination with scaling
Solve: 2x + 3y = 13 and 5x − y = 7
1
Multiply equation 2 by 3 to match the y coefficient: 15x − 3y = 21
2
Add to equation 1 (opposite signs on y, so add): 2x + 3y + 15x − 3y = 13 + 21 → 17x = 34 → x = 2
3
Substitute x = 2 into equation 2: 5(2) − y = 7 → y = 3
4
Check in equation 1: 2(2) + 3(3) = 4 + 9 = 13 ✓
Answerx = 2, y = 3
Substitution method
Rearrange one equation to make x or y the subject, then substitute that expression into the other equation. This method is particularly useful when one equation already has x or y isolated, and is essential when one equation is quadratic.
Worked Example 3 — Substitution (two linear equations)
Solve: y = 2x − 1 and 3x + 4y = 18
1
Equation 1 already gives y in terms of x. Substitute y = 2x − 1 into equation 2: 3x + 4(2x − 1) = 18
2
Expand and simplify: 3x + 8x − 4 = 18 → 11x = 22 → x = 2
3
Substitute back: y = 2(2) − 1 = y = 3
4
Check in equation 2: 3(2) + 4(3) = 6 + 12 = 18 ✓
Answerx = 2, y = 3
Worked Example 4 — Substitution (one linear, one quadratic)
Solve: y = x + 1 and x² + y² = 25
1
Substitute y = x + 1 into x² + y² = 25: x² + (x + 1)² = 25
2
Expand: x² + x² + 2x + 1 = 25 → 2x² + 2x − 24 = 0 → x² + x − 12 = 0
3
Factorise: (x + 4)(x − 3) = 0 → x = −4 or x = 3
4
Find y: when x = −4 → y = −4 + 1 = −3; when x = 3 → y = 3 + 1 = 4
5
Check: (−4)² + (−3)² = 16 + 9 = 25 ✓ 3² + 4² = 9 + 16 = 25 ✓
Answerx = −4, y = −3 or x = 3, y = 4
Graphical method
Draw both equations as graphs on the same axes. The coordinates of the intersection point(s) are the solution(s). This method gives approximate answers so accurate drawing is essential.
For two straight lines there will be exactly one intersection point (unless they are parallel). For one linear and one quadratic there can be zero, one or two intersection points.
Practice questions
Work through each question before checking the answers. Difficulty is shown for each question.
Foundation (Grade 3–5)
Q1Solve by elimination: x + y = 8 and x − y = 2Foundation
Q2Solve by elimination: 2x + y = 11 and 2x − y = 3Foundation
Q3Solve by substitution: y = x + 4 and x + y = 10Foundation
Q4Solve by elimination: 3x + 2y = 12 and x + 2y = 8Foundation
Q5Two numbers add to 20 and their difference is 6. Write two equations and find both numbers.Foundation
Higher (Grade 5–7)
Q6Solve: 3x + 4y = 25 and 5x − 2y = 7Higher
Q7Solve: 4x + 3y = 11 and 2x − 5y = −1Higher
Q8Solve by substitution: y = 3x − 2 and 4x + 2y = 20Higher
Q9A cinema sells adult tickets for £9 and child tickets for £5. In one session 80 tickets were sold for £560. How many of each type were sold?Higher
Q10Solve: 7x − 3y = 2 and 3x + y = 10Higher
Higher — Hard (Grade 8–9)
Q11Solve: y = x + 1 and x² + y² = 25Grade 8–9
Q12Solve: y = x + 2 and y = x² − 4Grade 8–9
Q13Solve: x² + y = 5 and x + y = 3Grade 8–9
Q14Find the coordinates of the points where the line y = x + 2 intersects the circle x² + y² = 20Grade 8–9
Q15Show algebraically that the line y = x + 3 does not intersect the curve y = x² + x + 5Grade 8–9
Answers
Foundation (Q1–Q5)
Q1x = 5, y = 3(add: 2x = 10 → x = 5; subtract: 2y = 6 → y = 3)
Q2x = 3.5, y = 4(add: 4x = 14 → x = 3.5; sub into eq1: 7 + y = 11 → y = 4)
Q3x = 3, y = 7(sub y = x + 4: x + x + 4 = 10 → 2x = 6 → x = 3; y = 7)
Q4x = 2, y = 3(subtract: 2x = 4 → x = 2; sub: 2 + 2y = 8 → y = 3)
Q513 and 7(x + y = 20 and x − y = 6; add: 2x = 26 → x = 13; y = 7)
Higher (Q6–Q10)
Q6x = 3, y = 4(multiply eq2 by 2: 10x − 4y = 14; add to eq1: 13x = 39 → x = 3; y = 4)
Q7x = 2, y = 1(multiply eq2 by 2: 4x − 10y = −2; subtract from eq1: 13y = 13 → y = 1; x = 2)
Q8x = 2.4, y = 5.2(4x + 2(3x − 2) = 20 → 10x = 24 → x = 2.4; y = 3(2.4) − 2 = 5.2)
Q940 adult, 40 child(a + c = 80; 9a + 5c = 560; multiply first eq by 5: subtract to get 4a = 160 → a = 40; c = 40)
Q10x = 2, y = 4(multiply eq2 by 3: 9x + 3y = 30; add to eq1: 16x = 32 → x = 2; y = 4)
Higher — Hard (Q11–Q15)
Q11x = 3, y = 4 or x = −4, y = −3(sub y=x+1: x²+(x+1)²=25 → 2x²+2x−24=0 → x²+x−12=0 → (x−3)(x+4)=0)
Q12x = 3, y = 5 or x = −2, y = 0(set equal: x+2=x²−4 → x²−x−6=0 → (x−3)(x+2)=0)
Q13x = 2, y = 1 or x = −1, y = 4(sub y=3−x: x²+3−x=5 → x²−x−2=0 → (x−2)(x+1)=0)
Q14x = 2, y = 4 or x = −4, y = −2(sub y=x+2: x²+(x+2)²=20 → 2x²+4x−16=0 → x²+2x−8=0 → (x−2)(x+4)=0)
Q15Discriminant = −8 < 0, so no real solutions — the line does not intersect the curve(set equal: x+3=x²+x+5 → x²+2=0; discriminant = 0−4(1)(2) = −8 < 0)
Common mistakes
These are the errors Alamin sees most frequently with simultaneous equations at GCSE. Recognising them now will save you marks in the exam.
Common Mistake 1
Adding when you should subtract (or vice versa)
If the coefficients you are eliminating have the same sign, you subtract the equations. If they have opposite signs, you add. Mixing these up gives a completely wrong answer. Always check the signs before deciding.
Common Mistake 2
Not multiplying every term when scaling
When multiplying an equation to match coefficients, every term must be multiplied — including the right-hand side constant. For example, multiplying 2x − y = 5 by 3 gives 6x − 3y = 15, not 6x − 3y = 5.
Common Mistake 3
Only finding one value and forgetting the other
After eliminating one variable and finding x, students sometimes forget to substitute back to find y. A simultaneous equation answer is always a pair of values — both x and y are required for full marks.
Common Mistake 4
Not expanding brackets correctly in substitution
When substituting y = 2x − 1 into 3x + 4y, students write 3x + 4 × 2x − 1 = 3x + 8x − 1 instead of 3x + 4(2x − 1) = 3x + 8x − 4. Always use brackets when substituting an expression.
Common Mistake 5
Only finding one solution for linear-quadratic pairs
When one equation is quadratic, the factorised equation usually gives two values of x, and therefore two pairs of solutions. Students frequently find one value and stop. Always check whether the quadratic has two roots and find both corresponding y values.
Common Mistake 6
Not checking the answer
Simultaneous equations are one of the few topics where checking is quick and definitive — substitute both values into both original equations. If either equation doesn't balance, you have made an error. This check costs 30 seconds and can save 2–3 marks.
Want to improve your grade faster?
If simultaneous equations are still causing problems, Alamin's diagnostic approach identifies exactly which skills are missing and builds a targeted plan to address them — with AI-powered practice between sessions to reinforce what's been taught.
Book an Assessment Session (£60)
No upfront payment required — payment is taken after confirmation.