Simultaneous Equations Calculator
Solve a system of two linear simultaneous equations instantly. Enter the coefficients for your two equations in the form ax + by = c and get step-by-step solutions using both the elimination and substitution methods. Aligned with GCSE and A-Level maths in England.
Enter your equations in the form ax + by = c. Use negative numbers for negative coefficients.
Equation 1:
Equation 2:
What Are Simultaneous Equations?
Simultaneous equations (also called a system of equations) are two or more equations that contain the same variables and must all be true at the same time. In UK GCSE and A-Level mathematics, you will most commonly encounter linear simultaneous equations — equations where the highest power of each variable is 1, giving straight lines when plotted on a graph.
The solution to a system of two linear equations in x and y is the point where their graphs intersect. Algebraically, it is the pair of values (x, y) that satisfies both equations simultaneously. For example:
- Equation 1: 2x + y = 7
- Equation 2: x − y = 2
- Solution: x = 3, y = 1 (verify: 2(3)+1 = 7 and 3−1 = 2)
Method 1: The Elimination Method
The elimination method is the most widely taught approach in UK GCSE maths. The goal is to manipulate the equations so that one variable cancels out when you add or subtract the equations.
Step-by-step elimination example:
Solve: 3x + 2y = 16 and 5x − 2y = 0
- The y coefficients are already equal in magnitude but opposite in sign (+2 and −2).
- Add the two equations: (3x + 5x) + (2y − 2y) = 16 + 0 → 8x = 16
- Solve for x: x = 16 / 8 = 2
- Substitute x = 2 back into Equation 1: 3(2) + 2y = 16 → 6 + 2y = 16 → 2y = 10 → y = 5
- Verify: 5(2) − 2(5) = 10 − 10 = 0 ✓
When the coefficients are not equal, multiply one or both equations by appropriate constants to make one set of coefficients match, then eliminate.
Method 2: The Substitution Method
Substitution is particularly efficient when one equation has a coefficient of 1 for one of the variables.
Step-by-step substitution example:
Solve: y = 2x − 1 and 3x + y = 9
- Equation 1 is already in the form y = 2x − 1 (y is the subject).
- Substitute into Equation 2: 3x + (2x − 1) = 9
- Simplify: 5x − 1 = 9 → 5x = 10 → x = 2
- Back-substitute: y = 2(2) − 1 = 3
- Verify: 3(2) + 3 = 6 + 3 = 9 ✓
Cramer’s Rule — The Matrix Approach
For the system a&sub1;x + b&sub1;y = c&sub1; and a&sub2;x + b&sub2;y = c&sub2;, Cramer’s Rule gives a direct formula using determinants:
The denominator D = a&sub1;b&sub2; − a&sub2;b&sub1; is the determinant. If D = 0, the system has no unique solution.
This calculator uses Cramer’s Rule internally to compute solutions. The method is taught at A-Level and forms the foundation for solving larger systems using matrices and Gaussian elimination.
Types of Solutions
| Outcome | Determinant D | Graph Interpretation | Number of Solutions |
|---|---|---|---|
| Unique solution | D ≠ 0 | Lines intersect at one point | Exactly 1 |
| No solution | D = 0, lines parallel | Lines never meet | 0 (inconsistent) |
| Infinite solutions | D = 0, same line | Lines are identical | Infinitely many (dependent) |
GCSE Exam Approach: Word Problems
On UK GCSE papers, simultaneous equations are often presented as word problems. The key skill is translating the English description into two algebraic equations. Here is a classic example type:
Problem: “Three pens and two rulers cost £2.60. One pen and four rulers cost £2.00. Find the cost of one pen and one ruler.”
Solution:
- Let p = price of a pen (in pence), r = price of a ruler (in pence).
- Equation 1: 3p + 2r = 260
- Equation 2: p + 4r = 200
- From Equation 2: p = 200 − 4r. Substitute into Equation 1: 3(200 − 4r) + 2r = 260
- 600 − 12r + 2r = 260 → −10r = −340 → r = 34p
- p = 200 − 4(34) = 200 − 136 = 64p
- Pen costs 64p, ruler costs 34p.
Simultaneous Equations with Quadratics (A-Level)
At A-Level, you will encounter systems where one equation is linear and one is quadratic. The technique is substitution from the linear into the quadratic, producing a quadratic in one variable. The discriminant then determines the number of intersection points (0, 1, or 2).
Example: Solve y = x + 3 and y = x² + 1.
- Substitute: x + 3 = x² + 1
- Rearrange: x² − x − 2 = 0
- Factorise: (x − 2)(x + 1) = 0 → x = 2 or x = −1
- y values: y = 2 + 3 = 5 and y = −1 + 3 = 2
- Solutions: (2, 5) and (−1, 2)
Graphical Method
Every linear equation represents a straight line on a coordinate grid. Plotting both lines and identifying the intersection point provides a visual solution. While less precise than algebraic methods (especially for non-integer solutions), graphing helps students understand why simultaneous equations have a unique solution in most cases.
To graph ax + by = c, find two points by setting x = 0 (find y-intercept) and y = 0 (find x-intercept), plot both points, and draw the line through them. Repeat for the second equation and find where the lines cross.
Real-World UK Applications
- Economics: Supply and demand equilibrium is found by solving two simultaneous equations — the supply function and the demand function. UK A-Level Economics students use this regularly.
- Catering and recipe scaling: Mixing two types of ingredients with different protein or calorie contents to meet nutritional targets requires simultaneous equation thinking.
- Currency exchange: Finding the break-even point when two exchange rate deals are compared involves setting up and solving simultaneous equations.
- Electrical circuits: Kirchhoff’s laws for circuit analysis produce simultaneous equations in the form of multiple voltage or current unknowns.
- Scheduling and logistics: Optimisation problems in UK business studies often reduce to simple systems of equations.
Frequently Asked Questions
What are simultaneous equations?
Simultaneous equations are two or more equations with the same variables that must all be satisfied at once. The solution is the set of variable values making all equations true simultaneously.
What is the best method to solve simultaneous equations?
Both elimination and substitution work for all linear systems. Elimination is generally faster when coefficients can be easily matched. Substitution is easiest when one variable has coefficient 1. On UK GCSE papers, either method earns full marks.
How do I know if simultaneous equations have no solution?
If the determinant D = a&sub1;b&sub2; − a&sub2;b&sub1; = 0 and the lines are parallel (but distinct), there is no solution. The ratios of coefficients are equal but not the constant ratio: a&sub1;/a&sub2; = b&sub1;/b&sub2; ≠ c&sub1;/c&sub2;.
Can this calculator solve three equations with three unknowns?
This calculator solves two linear equations with two unknowns. For three-variable systems, you would need a 3x3 system solver. At A-Level, three-variable systems are typically introduced via matrices and Gaussian elimination.
How do simultaneous equations appear on GCSE exams?
Simultaneous equations appear at both Foundation (paired linear equations) and Higher tier (including linear-quadratic systems). They may be presented algebraically or as word problems. Always verify your solution by substituting back into both original equations.
What does it mean if both variables cancel out?
If both variables cancel during elimination and you are left with a true statement (e.g., 0 = 0), the system has infinitely many solutions (the lines are identical). If you get a false statement (e.g., 0 = 5), there is no solution (the lines are parallel).