A 2×2 system has two parallel lines. How many solutions does it have?

Zero — parallel lines never intersect.

Systems of Linear Equations

On This Page

What Is a Linear Equation?
Writing a System in Matrix Form Ax = b
The Three Solution Types
Geometric Interpretation in 2D and 3D
Method 1, Substitution
Method 2: Elimination (by Hand)
Method 3, Gaussian Elimination (Matrix Method)
Method 4, Cramer's Rule
12 Fully Worked Examples
Common Mistakes
10-Question Practice Quiz

§ 01What Is a Linear Equation?

A linear equation is one in which every variable appears to exactly the first power, with no products of variables, no exponents other than 1, and no trigonometric, exponential, or logarithmic functions of the variables.

Linear

2x + 3y = 7 Both x and y appear to the first power. The graph is a straight line.

Linear

x − 4y + 2z = 0 Three variables, all to the first power. The graph is a plane in 3D.

Not Linear

x² + y = 5 x is squared — this is a parabola, not a line.

Not Linear

xy = 3 x and y are multiplied together — this is a hyperbola.

A system of linear equations is a collection of two or more linear equations sharing the same set of variables. The variables are unknowns we are trying to find. A solution to the system is a list of values, one for each variable, that satisfies every equation simultaneously.

A system of m equations in n unknowns

a₁₁x₁ + a₁₂x₂ + ··· + a_1nx_n = b₁ a₂₁x₁ + a₂₂x₂ + ··· + a_2nx_n = b₂ ⋮ a_m1x₁ + a_m2x₂ + ··· + a_mnx_n = b_m

Here a_ij are the coefficients (known constants), x_j are the unknowns, and b_i are the right-hand-side constants.

When all the b_i are zero, the system is called homogeneous. When at least one b_i is non-zero, it is non-homogeneous. Every homogeneous system has at least the trivial solution x₁ = x₂ = ··· = xₙ = 0.

§ 02Writing a System in Matrix Form Ax = b

Any system of linear equations can be written compactly as a single matrix equation Ax = b, where A is the coefficient matrix, x is the column vector of unknowns, and b is the column vector of constants.

Take the system: 2x − y + 3z = 10, x + 4y − z = 5, −3x + 2y + z = −2.

Matrix form Ax = b

2	−1	3
1	4	−1
−3	2	1

−2

The coefficient matrix A is m×n (m rows, n columns). The unknown vector x is n×1. The constants vector b is m×1. Matrix multiplication Ax produces exactly the left-hand sides of all m equations simultaneously.

The Augmented Matrix

For computation, we append b as an extra column to A, separated by a vertical bar, forming the augmented matrix [A | b]. This single object contains all the information in the system and is what we operate on during row reduction.

Augmented matrix [A | b] for the system above

2	−1	3	10
1	4	−1	5
−3	2	1	−2

Why matrix form matters. Writing the system as Ax = b opens up every tool of linear algebra: row reduction, matrix inverses, determinants, eigenvalues. It also makes it obvious whether a solution exists — the system is consistent if and only if b lies in the column space of A.

§ 03The Three Solution Types

Every system of linear equations falls into exactly one of three categories. There is no fourth option.

Type	Description	Terminology	Row Reduction Signature
Unique solution	Exactly one set of values satisfies all equations.	Consistent & determined	Every column of A has a pivot; no zero-row contradiction.
Infinitely many solutions	A whole family of solutions (a line, plane, etc.) satisfies the system.	Consistent & underdetermined	At least one non-pivot column (free variable); no contradiction row.
No solution	No set of values satisfies all equations simultaneously.	Inconsistent	A row of the form [0 0 ··· 0 \| c] with c ≠ 0 appears.

The only check you need: look for a contradiction row first. Before classifying a solution, always scan the row-reduced matrix for a row [0 0 ··· 0 | c] with c ≠ 0. If one exists, the system is inconsistent — full stop. Otherwise, count free variables: zero free variables means a unique solution; one or more free variables means infinitely many.

Consistent and Inconsistent

A system is consistent if it has at least one solution (either exactly one or infinitely many). It is inconsistent if it has no solution. This single binary distinction is often the first question asked about any system.

Overdetermined and Underdetermined

A system with more equations than unknowns (m > n) is overdetermined. It is usually (but not always) inconsistent. A system with fewer equations than unknowns (m < n) is underdetermined. If it is consistent, it automatically has infinitely many solutions because there will always be at least one free variable.

§ 04Geometric Interpretation in 2D and 3D

Linear algebra has a beautiful geometric face. Every linear equation in n variables describes a geometric object in ℝⁿ, and the solution of a system is the intersection of those objects.

Two Equations in Two Unknowns — Lines in ℝ²

Each equation ax + by = c describes a straight line in the xy-plane. Solving the system means finding all points (x, y) that lie on both lines simultaneously.

Case 1 — Unique solution

Two distinct lines cross at exactly one point. One unique solution.

Case 2 — Infinitely many

Two coincident lines (same equation scaled). Every point on the line is a solution.

Case 3 — No solution

Two parallel lines never meet. The system is inconsistent — no solution.

Three Equations in Three Unknowns — Planes in ℝ³

Each equation ax + by + cz = d describes a flat plane in three-dimensional space. Three planes can intersect in a variety of ways:

Unique solution

Three planes meet at a single point. Like the corner of a room where three walls meet — exactly one point satisfies all three equations.

Infinitely many — a line

Three planes share a common line. Two of the planes are consistent with the third but leave one free variable — the solution is parameterised by a line.

Infinitely many — a plane

All three planes are the same plane (or all contain a common plane). Two free variables; the solution is an entire plane.

No solution

The planes have no common intersection. Could be parallel planes, a "triangular prism" arrangement, or two planes intersecting in a line parallel to the third.

The geometric viewpoint is powerful. When you stare at an augmented matrix, you are really asking: "in what way do these hyperplanes intersect?" Understanding this geometry prevents mechanical errors — you know in advance roughly what kind of answer to expect.

§ 05Method 1 — Substitution

Substitution is the most elementary solving technique, best suited to small systems (2×2 or 3×3) where one variable is easy to isolate.

Step 1

Isolate one variable

Choose any equation and solve it for one variable in terms of the others. Choose the simplest isolation — avoid fractions where possible.

Step 2

Substitute into the other equations

Replace every occurrence of the isolated variable in all other equations with the expression found in Step 1. This reduces the system by one variable and one equation.

Step 3

Solve and back-substitute

Solve the reduced system, then substitute back to find the remaining variables. Always verify by plugging the solution into every original equation.

When to use substitution. Substitution shines when a variable already has coefficient 1 or −1 in some equation (isolating it produces no fractions). For larger systems or systems with no clean coefficients, Gaussian elimination on the augmented matrix is faster and less error-prone.

§ 06Method 2 — Elimination (by Hand)

Elimination (also called addition–elimination or the linear combination method) works by multiplying equations by constants and adding them together to cancel a variable.

Step 1

Choose a variable to eliminate

Pick one variable and a pair of equations. Multiply one or both equations so that the chosen variable has equal and opposite coefficients.

Step 2

Add to cancel

Add the two modified equations together. The target variable vanishes, leaving an equation in the remaining variables.

Step 3

Repeat and solve

Use different equation pairs to eliminate the same variable from all other equations, then solve the resulting smaller system.

Elimination by hand is equivalent to performing row operations on the augmented matrix: the two methods are identical in mathematics, just different in notation. For systems of 4 or more equations, the matrix version (Gaussian elimination) is strongly preferred because it is less prone to bookkeeping errors.

§ 07Method 3 — Gaussian Elimination (Matrix Method)

Gaussian elimination is the standard algorithmic method for solving any linear system. It works by transforming the augmented matrix [A | b] through row operations into Row Echelon Form (REF) or Reduced Row Echelon Form (RREF), from which the solution is directly readable. For a full treatment of the algorithm, see the Row Reduction and Echelon Form page.

The algorithm in four lines

1. Write the augmented matrix [A | b].

2. Apply row operations (swap, scale, add) to reach REF.

3. Check for a contradiction row [0 0 ··· 0 | c ≠ 0] → inconsistent if found.

4. Continue to RREF (or use back-substitution from REF) to read off the solution. Free variables parameterise the solution if any non-pivot columns exist.

Using the Matrix Inverse

For a square system (m = n), if the coefficient matrix A is invertible (det A ≠ 0), the unique solution is given directly by:

Solution via matrix inverse

x = A⁻¹b

This is exact but computationally expensive for large systems. In practice, row reduction is preferred because computing A⁻¹ requires roughly three times as much work as solving Ax = b directly.

§ 08Method 4 — Cramer's Rule

Cramer's Rule gives an explicit formula for each variable in a square, non-singular (invertible) system. It expresses each unknown as a ratio of two determinants.

Cramer's Rule for Ax = b, det(A) ≠ 0

x_j = det(A_j) / det(A)

where A_j is the matrix formed by replacing the j-th column of A with the vector b.

Cramer's Rule: beautiful but impractical for large systems. For a 2×2 or 3×3 system, Cramer's Rule is quick and elegant. For an n×n system with large n, computing n+1 determinants is far more expensive than Gaussian elimination. Use Cramer's Rule for small, symbolic problems and as a theoretical tool — not for numerical computation.

§ 0912 Fully Worked Examples

Example 012×2 — substitution, unique solution

System: y = 3x − 1 and 2x + y = 9.

Substitute

y is already isolated. Substitute y = 3x − 1 into equation 2: 2x + (3x − 1) = 9 → 5x = 10 → x = 2.

Back-sub

y = 3(2) − 1 = 5.

Verify

Eq 1: 5 = 3(2) − 1 = 5 ✓. Eq 2: 2(2) + 5 = 9 ✓.

Solution

x = 2, y = 5

Example 022×2 — elimination, unique solution

System: 3x + 2y = 12 and 5x − 2y = 4.

Eliminate y

The y-coefficients are already +2 and −2. Adding the two equations: (3x + 2y) + (5x − 2y) = 12 + 4 → 8x = 16 → x = 2.

Find y

Substitute into equation 1: 3(2) + 2y = 12 → 2y = 6 → y = 3.

Solution

x = 2, y = 3

Geometric meaning: the two lines cross at the point (2, 3).

Example 032×2 — inconsistent (parallel lines)

System: 2x + 4y = 6 and x + 2y = 5.

Row reduce

2	4	6
1	2	5

R₁↔R₂, then R₂→R₂−2R₁

1	2	5
0	0	−4

Row 2 reads 0 = −4. Contradiction.

Conclusion

Inconsistent — no solution. The lines are parallel (same slope, different intercepts).

Example 042×2 — infinitely many solutions (coincident lines)

System: 4x − 6y = 10 and −2x + 3y = −5.

Observe

Multiply equation 2 by −2: 4x − 6y = 10, which is identical to equation 1. The two equations are the same line.

Row reduce

2	−3	5
0	0	0

After scaling R₁ → (1/2)R₁: x − (3/2)y = 5/2. y is free (y = t).

General Solution

x = 5/2 + (3/2)t, y = t, t ∈ ℝ

Example 053×3 — Gaussian elimination, unique solution

System: x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2.

Augment

1	1	1	6
2	−1	1	3
1	2	−1	2

R₂→R₂−2R₁, R₃→R₃−R₁

1	1	1	6
0	−3	−1	−9
0	1	−2	−4

R₃→3R₃+R₂

1	1	1	6
0	−3	−1	−9
0	0	−7	−21

Back-sub: z = 3, y = (−9+3)/−3 = 2, x = 6−2−3 = 1.

Solution

x = 1, y = 2, z = 3

Example 063×3 — inconsistent system (triangular prism)

System: x + y = 3, y + z = 4, x + z = 2, x + y + z = 6.

Reduce

Adding all three two-variable equations: 2x + 2y + 2z = 9, so x + y + z = 4.5. But the fourth equation says x + y + z = 6. Since 4.5 ≠ 6, the system is inconsistent.

Conclusion

No solution — the four planes have no common intersection point.

Example 073×3 — one free variable (line of solutions)

System: x + y + 2z = 4, 2x + 2y + 4z = 8, 3x + 3y + 6z = 12.

Observe

Every equation is a multiple of the first: eq 2 = 2×eq 1, eq 3 = 3×eq 1. After row reduction, only one independent equation survives.

RREF

1	1	2	4
0	0	0	0
0	0	0	0

One pivot, two free variables (y = s, z = t). From row 1: x = 4 − s − 2t.

General Solution — a Plane

x = 4 − s − 2t, y = s, z = t, s,t ∈ ℝ

Example 08Cramer's Rule — 2×2 system

System: 3x + y = 7, x + 2y = 4. Apply Cramer's Rule.

Coefficient matrix

A = [[3, 1], [1, 2]], det(A) = 3·2 − 1·1 = 5

x: replace col 1 with b

A₁ = [[7, 1], [4, 2]], det(A₁) = 14 − 4 = 10 x = det(A₁)/det(A) = 10/5 = 2

y: replace col 2 with b

A₂ = [[3, 7], [1, 4]], det(A₂) = 12 − 7 = 5 y = det(A₂)/det(A) = 5/5 = 1

Solution

x = 2, y = 1

Example 09Matrix inverse method — 2×2

System: 4x + 7y = 29, 2x + 6y = 20. Solve using A⁻¹.

Find A⁻¹

det(A) = 4·6 − 7·2 = 10 A⁻¹ = (1/10)[[6, −7], [−2, 4]]

x = A⁻¹b

[x, y] = (1/10)[[6·29 + (−7)·20], [(−2)·29 + 4·20]] = (1/10)[[174 − 140], [−58 + 80]] = (1/10)[34, 22] = [3.4, 2.2]

Solution

x = 17/5, y = 11/5

Example 10Parametric system — find k for consistency

System: x + 3y − z = 2, 2x + y + z = 4, 3x + 4y + kz = 6. For which value of k is the system consistent?

Row reduce

1	3	−1	2
2	1	1	4
3	4	k	6

R₂ → R₂ − 2R₁: [0, −5, 3 | 0]. R₃ → R₃ − 3R₁: [0, −5, k+3 | 0].

R₃ → R₃ − R₂: [0, 0, k+3−3 | 0] = [0, 0, k | 0].

Interpret

The last row reads 0 = 0 for any k (since b = 0 in this row). So the system is consistent for all values of k. But the solution type changes: if k ≠ 0, unique solution; if k = 0, infinitely many (z is free).

Conclusion

Consistent for all k. Unique solution when k ≠ 0; infinitely many when k = 0.

Example 11Real-world application — traffic flow

Cars flow into a junction network. At each junction, cars in = cars out (conservation). The flows x₁, x₂, x₃ on three road segments satisfy:

Flow equations

x₁ − x₂ = 10 (junction A) x₂ + x₃ = 30 (junction B) x₁ + x₃ = 40 (junction C)

Row reduce

1	−1	0	10
0	1	1	30
1	0	1	40

R₃→R₃−R₁

1	−1	0	10
0	1	1	30
0	1	1	30

R₃ → R₃ − R₂ gives a zero row. x₃ is free.

Solution — traffic can flow in many valid patterns

x₁ = 40 − t, x₂ = 30 − t, x₃ = t, 0 ≤ t ≤ 30

The free parameter t represents the flow on segment 3. Any value between 0 and 30 gives a valid, non-negative traffic assignment.

Example 124×4 system — full Gaussian elimination

System: w + x + y + z = 10, 2w − x + y − z = 2, w + 3x − y + z = 8, w − x + 2y + 2z = 12.

Augment & Reduce

1	1	1	1	10
2	−1	1	−1	2
1	3	−1	1	8
1	−1	2	2	12

R₂ → R₂ − 2R₁, R₃ → R₃ − R₁, R₄ → R₄ − R₁:

1	1	1	1	10
0	−3	−1	−3	−18
0	2	−2	0	−2
0	−2	1	1	2

Continue: R₃ → 3R₃ + 2R₂ gives [0, 0, −8, −6 | −42]. R₄ → 3R₄ − 2R₂ gives [0, 0, 5, 9 | 42].

Then eliminate the last variable: 8R₄+5R₃: [0,0,0,42|126] → z = 3. Back-sub: y = 42/8 − ... continuing: z = 3, y = 2, x = 3, w = 2.

Solution

w = 2, x = 3, y = 2, z = 3

§ 10Common Mistakes

Mistake 1 — Forgetting to check for a contradiction before writing the solution. Always inspect the fully reduced matrix for a row [0 0 ··· 0 | c ≠ 0] before concluding the system is consistent. Many students skip this and write a "solution" for an inconsistent system.

Mistake 2 — Dropping the right-hand side during row operations. Every row operation applies to the entire augmented row, including the constant column after the bar. Applying an operation to [A] but not to [b] gives a different system with different solutions.

Mistake 3 — Setting free variables to zero without justification. When a free variable exists, the general solution must parameterise over all its values. Setting it to zero gives only one particular solution — the problem asks for the complete solution set.

Mistake 4 — Confusing "no solution" with "infinitely many solutions". An all-zero row [0 0 ··· 0 | 0] is not a contradiction — it just disappears and signals a free variable (infinitely many solutions). The contradiction comes only when the constant is non-zero: [0 0 ··· 0 | c ≠ 0].

Good habit — always verify by substituting the solution back into every original equation. A 30-second check catches arithmetic errors that might otherwise cost full marks. For parameterised solutions, verify that the formula works for at least two different values of the parameter.

§ 11Practice Quiz

Test your understanding of linear systems, solution types, geometric meaning, methods, and special cases. Instant feedback on every question.

Systems of Linear Equations

10 Questions