Substitution, elimination, Gaussian elimination, matrices, and 3-variable systems — with worked examples.
| Method | Best For | Key Steps |
|---|---|---|
| Substitution | 2 variables, one easily isolated | Solve for one variable → plug into other equation → solve → back-substitute |
| Elimination | 2 variables, integer coefficients | Multiply equations to match a coefficient → add/subtract to eliminate → solve → back-substitute |
| Gaussian Elimination | 3+ variables, any system | Write augmented matrix → row operations → row echelon form → back-substitute |
| Cramer's Rule | 2–3 variables with determinants | Compute det(A), replace each column with constants, divide |
Solve one equation for one variable, then substitute the expression into the other equation.
STEPS
Example
Solve: x + 2y = 8 and 3x − y = 3
Step 1: Solve equation 1 for x: x = 8 − 2y
Step 2: Substitute: 3(8 − 2y) − y = 3
Step 3: 24 − 6y − y = 3 → −7y = −21 → y = 3
Step 4: x = 8 − 2(3) = 2
Solution: (2, 3) ✓
Multiply equations so that one variable has opposite coefficients, then add equations to eliminate it.
Example
Solve: 2x + 3y = 12 and 4x − y = 10
Multiply eq2 by 3: 12x − 3y = 30
Add equations: (2x + 3y) + (12x − 3y) = 12 + 30
14x = 42 → x = 3
Back-sub: 2(3) + 3y = 12 → 3y = 6 → y = 2
Solution: (3, 2) ✓
Recognizing Special Cases
No Solution (Inconsistent)
After eliminating, you get: 0 = 5 (contradiction)
Lines are parallel — never intersect
Infinite Solutions (Dependent)
After eliminating, you get: 0 = 0 (identity)
Same line — write as parametric solution
Write the system as an augmented matrix and apply row operations to reach row echelon form.
| Operation | Notation | Effect |
|---|---|---|
| Swap rows | R₁ ↔ R₂ | Exchange two rows |
| Scale a row | kRᵢ → Rᵢ | Multiply every entry in a row by k ≠ 0 |
| Row combination | Rᵢ + kRⱼ → Rᵢ | Add k times row j to row i |
3-Variable Example
System: x + y + z = 6 | 2x − y + z = 3 | x + 2y − z = 2
Augmented matrix: [1 1 1 | 6] [2 −1 1 | 3] [1 2 −1 | 2]
R₂ − 2R₁ → R₂: [0 −3 −1 | −9]
R₃ − R₁ → R₃: [0 1 −2 | −4]
Swap R₂, R₃: [0 1 −2 | −4] [0 −3 −1 | −9]
R₃ + 3R₂ → R₃: [0 0 −7 | −21] → z = 3
Back-sub: y − 2(3) = −4 → y = 2 | x + 2 + 3 = 6 → x = 1
Solution: (1, 2, 3) ✓
For a 2×2 system ax + by = e and cx + dy = f:
det(A) = ad − bc
x = (ed − bf) / det(A)
y = (af − ce) / det(A)
Replace the x-column with constants for Dₓ, y-column for Dᵧ
When Cramer's Rule Fails
If det(A) = 0, the system is either inconsistent (no solution) or dependent (infinite solutions). Cramer's rule does not apply — use elimination to determine which case.
Systems where at least one equation is not linear (contains x², xy, circles, parabolas, etc.). Use substitution.
Example: Line and Circle
x² + y² = 25 (circle) and y = x + 1 (line)
Substitute y = x + 1: x² + (x+1)² = 25
x² + x² + 2x + 1 = 25 → 2x² + 2x − 24 = 0
x² + x − 12 = 0 → (x+4)(x−3) = 0
x = −4 or x = 3 → y = −3 or y = 4
Solutions: (3, 4) and (−4, −3) ✓ — two intersection points
A line can intersect a conic in 0, 1, or 2 points. Two conics can intersect in up to 4 points.
Exactly one solution. Lines (or planes) intersect at exactly one point. det(A) ≠ 0.
No solution. Lines are parallel (same slope, different intercept). Contradiction appears during solving.
Infinitely many solutions. Same line (equations are multiples). Identity 0 = 0 appears during solving.
Leading entry of each row is 1 (pivot), each pivot is to the right of the pivot above it, zeros below each pivot.
|a b; c d| = ad − bc. Zero determinant means no unique solution.
Expand along first row: a(ei−fh) − b(di−fg) + c(dh−eg)
The three main methods are: (1) Substitution — solve one equation for a variable and plug into the other; (2) Elimination (linear combination) — add or subtract equations to eliminate a variable; (3) Matrices — write the system as an augmented matrix and use row operations (Gaussian elimination) or Cramer's rule. Substitution is best for simple systems; matrices are best for 3+ variable systems.
Use Gaussian elimination: write the system as a 3×4 augmented matrix, then use row operations (swap rows, multiply a row by a scalar, add a multiple of one row to another) to get row echelon form (upper triangular). Then back-substitute to find the variables. Alternatively, use Cramer's rule with a 3×3 determinant.
No solution (inconsistent system) occurs when the equations represent parallel lines (or parallel planes in 3D) — you get a contradiction like 0 = 5 during solving. Infinitely many solutions (dependent system) occurs when equations represent the same line/plane — you get 0 = 0. A unique solution is when the lines/planes intersect at exactly one point.
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