Systems of Equations — Complete Guide
A system of equations is a set of two or more equations that share the same unknowns. The solution is the set of values that satisfies all equations simultaneously. Systems arise everywhere — from finding the break-even point of a business to calculating the forces in a bridge to balancing chemical equations.
This guide covers every standard method for solving 2×2 linear systems: substitution, elimination (addition/subtraction), Cramer's rule using determinants, and graphical interpretation. A final section extends the methods to 3×3 systems.
Types of Systems — Consistent, Inconsistent, Dependent
Every system of two linear equations in two unknowns falls into one of three categories, determined by the geometric relationship of the two lines:
| Type | Lines | Solutions | Determinant D | Example |
|---|---|---|---|---|
| Consistent & independent | Intersect at one point | Exactly one | D ≠ 0 | x+y=5, x−y=1 |
| Inconsistent | Parallel (never meet) | None | D = 0 | x+y=3, x+y=7 |
| Consistent & dependent | Identical (same line) | Infinitely many | D = 0 | x+y=3, 2x+2y=6 |
Method 1: Substitution
Substitution works by solving one equation for one variable and substituting that expression into the other equation.
Steps: (1) Solve one equation for one variable. (2) Substitute that expression into the other equation. (3) Solve the resulting single-variable equation. (4) Substitute back to find the other variable. (5) Verify in both original equations.
Example: Solve: x + 2y = 7 and 3x − y = 4.
Step 1: From eq.1, x = 7 − 2y.
Step 2: Substitute: 3(7−2y) − y = 4 → 21 − 6y − y = 4 → 21 − 7y = 4.
Step 3: −7y = −17 → y = 17/7 … hmm. Let's use a cleaner example.
Clean example: x + y = 10 and x − y = 4. From eq.1: x = 10 − y. Sub into eq.2: (10−y) − y = 4 → 10 − 2y = 4 → y = 3. Then x = 10 − 3 = 7. Solution: (7, 3). Verify: 7+3=10 ✓, 7−3=4 ✓.
Best use of substitution: When one equation already has a variable with coefficient 1 or −1, substitution is fast. If all coefficients are large, prefer elimination.
Method 2: Elimination (Addition/Subtraction)
Elimination works by multiplying equations by constants to make the coefficients of one variable opposite in sign, then adding the equations to eliminate that variable.
Steps: (1) Choose a variable to eliminate. (2) Multiply one or both equations so that variable has equal and opposite coefficients. (3) Add the equations. (4) Solve the resulting single-variable equation. (5) Back-substitute to find the other variable.
Example: Solve: 2x + 3y = 12 and 5x − y = 7.
To eliminate y: multiply eq.2 by 3 → 15x − 3y = 21. Add eq.1: (2x + 3y) + (15x − 3y) = 12 + 21 → 17x = 33 → x = 33/17. Back-sub: 2(33/17) + 3y = 12 → 3y = 12 − 66/17 = 138/17 → y = 46/17.
Cleaner example: x + y = 7 and x − y = 3. Add: 2x = 10 → x = 5. Subtract eq.2 from eq.1: 2y = 4 → y = 2. Solution (5, 2). Verify: 5+2=7 ✓, 5−2=3 ✓.
Best use of elimination: When coefficients have an obvious LCM, or when two equations have the same variable coefficient (subtract immediately).
Method 3: Cramer's Rule (Determinants)
Cramer's Rule uses determinants to solve a system directly from coefficients. For the system a₁x + b₁y = c₁ and a₂x + b₂y = c₂:
Step 1: Compute D = a₁b₂ − a₂b₁ (the coefficient determinant).
Step 2: If D = 0, the system is either inconsistent or dependent (no unique solution).
Step 3: Compute Dₓ = c₁b₂ − c₂b₁ (replace a-column with c-values).
Step 4: Compute Dᵧ = a₁c₂ − a₂c₁ (replace b-column with c-values).
Step 5: x = Dₓ/D, y = Dᵧ/D.
Example: 3x + 2y = 8 and x − y = 1. D = (3)(−1) − (1)(2) = −3 − 2 = −5. Dₓ = (8)(−1) − (1)(2) = −8 − 2 = −10. Dᵧ = (3)(1) − (1)(8) = 3 − 8 = −5. x = −10/−5 = 2, y = −5/−5 = 1. Solution (2,1). Cramer's Rule is especially useful when you need only one variable's value, or when coding numerical methods.
No Solution and Infinite Solutions — Recognising Them
No solution (inconsistent): When you reach a contradiction like 0 = 5 or 3 = 7 during solving, the system has no solution. The lines are parallel. Example: x + y = 3 and x + y = 7. Subtracting: 0 = 4 — contradiction. No solution.
Infinite solutions (dependent): When you reach an identity like 0 = 0 during solving, the two equations are multiples of each other — the same line. Any point on that line is a solution. Express as {(x, y) | x + y = 3} or parametrically: x = t, y = 3 − t for all real t.
Geometric interpretation: Parallel lines never meet (no solution). Identical lines overlap everywhere (infinite solutions). Only intersecting lines produce exactly one solution.
3×3 Systems — Row Reduction Overview
Systems with 3 equations and 3 unknowns are solved by the same principles — substitute or eliminate to reduce to 2×2, then solve.
Elimination for 3×3: Choose a variable to eliminate first. Use equation pairs to eliminate that variable from 2 of the 3 equations, leaving a 2×2 system in the remaining two variables. Solve the 2×2 system, then back-substitute.
Example system: x+y+z=6, 2x−y+z=3, x+2y−z=2.
Eliminate z from eq.1 and eq.3: eq.1+eq.3 → 2x+3y=8. Eliminate z from eq.2 and eq.1: eq.2−eq.1 → x−2y=−3. Solve 2×2: multiply second by 2: 2x−4y=−6. Subtract from first: 7y=14 → y=2. Then x = −3+2(2) = 1. Then z = 6−1−2 = 3. Solution: (1, 2, 3).
Gauss-Jordan row reduction is the systematic method for larger systems — it uses elementary row operations on the augmented matrix to reach reduced row echelon form. Most calculators and linear algebra software use this approach.
Real-World Applications of Systems of Equations
Break-even analysis: Revenue R = px (price × units) and Cost C = F + vx (fixed cost + variable cost × units). Set R = C and solve for x (break-even quantity).
Mixture problems: Mixing two solutions of different concentrations. If you mix x litres of 20% acid with y litres of 50% acid to get 10 litres of 35% acid: x + y = 10 and 0.2x + 0.5y = 3.5. Solve by substitution or elimination.
Distance/rate/time: A boat travels 30 km downstream in 2 hours and 30 km upstream in 3 hours. If b = boat speed and c = current: b + c = 15 and b − c = 10. Solve: b = 12.5, c = 2.5 km/h.
Investment problems: Two investments totalling $10,000 earning 5% and 8% yield $710 total. x + y = 10,000 and 0.05x + 0.08y = 710. Solve to find each investment amount.