Four Methods to Solve Simultaneous Equations

All methods start with the same matrix equation Ax = b. Each exploits different properties of the matrix — choose the method the question requires, or the one that suits the system size.

Determinant Method Cramer's Rule Inverse Matrix Gaussian Elimination 2×2 & 3×3 Systems

📐

Determinant Method

Find det(A). If zero — no solution. Essential foundation for Cramer's Rule.

Foundation

🧮

Cramer's Rule

Replace each column with b, divide determinant ratios. Step-by-step animated.

Determinants

🔄

Inverse Method

Find A⁻¹ then compute x = A⁻¹b. Cofactors highlighted live.

Matrix Inverse

📊

Gaussian Elimination

Row operations on [A|b] animated step by step. Best for any size system.

Row Operations

📐The Matrix Equation Ax = bFoundation

Every simultaneous system can be written as one matrix equation. For two equations:

a₁x + b₁y = c₁ → [a₁ b₁] [x] [c₁] a₂x + b₂y = c₂ [a₂ b₂] [y] = [c₂] A x = b

For three equations (3×3 system):

a₁x + b₁y + c₁z = d₁ → [a₁ b₁ c₁] [x] [d₁] a₂x + b₂y + c₂z = d₂ [a₂ b₂ c₂] [y] = [d₂] a₃x + b₃y + c₃z = d₃ [a₃ b₃ c₃] [z] [d₃]

🎯 All four methods operate on exactly this form. The difference is only in how they use matrix A to find x.

📐

Determinant Method

Solve simultaneous equations using determinants — calculate D, Dx and Dy, then divide to find each unknown.

🎨 Colour Key — the same colours are used everywhere on this page

a₁, a₂

x-coefficients
Column 1 of D

b₁, b₂

y-coefficients
Column 2 of D

c₁, c₂

Constants
(right-hand side)

D Dx Dy

The three
determinants

💡 Subscript 1 = equation 1 | Subscript 2 = equation 2 | Green cells always show where constants have been swapped in

🔢Reading the EquationsWhat each letter means

Every number in the equations has a label. Here's what each label means:

📘 Equation 1

coeff of x
eq 1

coeff of y
eq 1

constant
eq 1

📗 Equation 2

coeff of x
eq 2

coeff of y
eq 2

constant
eq 2

📊How D, Dx and Dy Are FormedVisual Guide

Three 2×2 grids. Colours show exactly which numbers go where. Green = constants swapped in.

D

Main Determinant

a₁

b₁

a₂

b₂

x-coefficients in col 1
y-coefficients in col 2
No constants here

a₁b₂ − b₁a₂

Dx

col 1 → constants

c₁

b₁

c₂

b₂

Constants c₁,c₂ replace col 1
y-coefficients stay in col 2

c₁b₂ − b₁c₂

Dy

col 2 → constants

a₁

c₁

a₂

c₂

x-coefficients stay in col 1
Constants c₁,c₂ replace col 2

a₁c₂ − c₁a₂

Find x

x = Dx ÷ D

Find y

y = Dy ÷ D

⚠️ If D = 0 — stop! No unique solution. You cannot divide by zero.

✏️Worked ExampleFull walkthrough

Solve these simultaneous equations

D = det(A)

1

1

1

−1

(1×−1) − (1×1)
= −1 − 1 = −2

Dx

5

1

7

−1

(5×−1) − (1×7)
= −5 − 7 = −12

Dy

1

5

1

7

(1×7) − (5×1)
= 7 − 5 = 2

x = Dx ÷ D

−12 ÷ −2 = 6

y = Dy ÷ D

2 ÷ −2 = −1

✅ x = 6 y = −1
Check eq1: 6 + (−1) = 5 ✓ Check eq2: 6 − (−1) = 7 ✓

🎛️Interactive Solver — try your own equationsLive

Enter your coefficients and constants — colours match the guide above. D, Dx and Dy update as you type.

📘 Equation 1: a₁x + b₁y = c₁

x + y =

📗 Equation 2: a₂x + b₂y = c₂

x + y =

👆Try 2×2 live: enter x+y=5, x−y=7 — D, Dx, Dy update as you type · Switch to 3×3 and step through D, Dx, Dy, Dz · Try det=0: 2x+4y=6, x+2y=3 — system has no unique solution.

🧮

Cramer's Rule

Each unknown is a ratio of two determinants. Replace the column for that unknown with the constants vector, then divide by D.

🎨 Same colour system as the Determinant Method tab

a₁, a₂

x-coefficients

b₁, b₂

y-coefficients

c₁, c₂

constants (b vector)

D Dx Dy

the determinants

💡 Green cells always show where constants have been swapped into the matrix column.

💡What Is Cramer's Rule?Core Idea

Cramer's Rule says: to find x, replace the x-column of the coefficient matrix with the constants, find its determinant, then divide by D. Same idea for y.

D

Coefficient matrix
No constants
x-coeffs + y-coeffs only

Dx

Replace col 1
(the x-column)
with constants

Dy

Replace col 2
(the y-column)
with constants

Find x

x = Dx ÷ D

Find y

y = Dy ÷ D

⚠️ If D = 0 — no unique solution. Cramer's Rule cannot be applied.

📋Step-by-Step ProcedureMethod

Rearrange — all variable terms on the left, constants on the right.
Identify — label your a₁, a₂ (x-coeffs), b₁, b₂ (y-coeffs) and c₁, c₂ (constants).
Calculate D = a₁b₂ − b₁a₂. If D = 0, stop — no unique solution.
Form Dx — swap col 1 for c₁, c₂. Calculate Dx = c₁b₂ − b₁c₂.
Form Dy — swap col 2 for c₁, c₂. Calculate Dy = a₁c₂ − c₁a₂.
Divide — x = Dx÷D, y = Dy÷D.
Check — substitute x and y back into both original equations.

✏️Worked Example — 2×2Full walkthrough

Solve using Cramer's Rule

        3x +
        5y =
        7
         (Eq 1: a₁=3, b₁=5, c₁=7)

        4x
        −3y =
        19
         (Eq 2: a₂=4, b₂=−3, c₂=19)
      

D = det(A)

coefficient matrix

3

5

4

−3

(3×−3) − (5×4)
= −9 − 20
= −29

Dx

col 1 → constants

7

5

19

−3

(7×−3) − (5×19)
= −21 − 95
= −116

Dy

col 2 → constants

3

7

4

19

(3×19) − (7×4)
= 57 − 28
= 29

x = Dx ÷ D

−116 ÷ −29 = 4

y = Dy ÷ D

29 ÷ −29 = −1

✅ x = 4 y = −1
Check eq1: 3(4)+5(−1) = 12−5 = 7 ✓ Check eq2: 4(4)−3(−1) = 16+3 = 19 ✓

🎛️Visual Step-by-Step SolverInteractive

Enter your system. Red inputs = x-coefficients | Yellow inputs = y-coefficients | Green inputs = constants

📘 Equation 1

x + y =

📗 Equation 2

x + y =

–

Ready

Enter your equations above then press ▶ Show Step-by-Step

Each step shows which column turns green as the constants are swapped in, with the determinant value displayed alongside.

🔄

Inverse Matrix Method

Find A⁻¹ then multiply by b — x = A⁻¹b gives all unknowns at once.

🎨 Same colour system throughout this page

a, c

x-coefficients
(column 1 of A)

b, d

y-coefficients
(column 2 of A)

e, f

constants
(vector b)

A⁻¹

inverse matrix
(shown in green)

💡 Orange = swapped diagonal elements | Purple = negated off-diagonal | Green = final A⁻¹ values

💡The Core IdeaTheory

Start from Ax = b. Multiply both sides on the left by A⁻¹:

Ax = b

↓ multiply both sides by A⁻¹

A⁻¹Ax = A⁻¹b

↓ A⁻¹A = I (identity)

x = A⁻¹b ✅

For A = [a b; c d], the inverse formula is:

Step 1 — Find det(A)

det = ad − bc

Step 2 — Build A⁻¹

↘ Swap a↔d

d   −b

↗ Negate b&c

−c   a

Then divide every element by det

✏️Worked Example — 2×2Full walkthrough

Solve using the Inverse Method

        3x +
        5y =
        7
        (a=3, b=5, e=7)

        4x
        −3y =
        19
        (c=4, d=−3, f=19)
      

Step 1 — Matrix A

3

5

4

−3

det = (3×−3)−(5×4)
= −29

Step 2 — Swap & Negate

−3

−5

−4

3

Orange=diagonal swapped
Purple=off-diag negated

Step 3 — A⁻¹ (÷ −29)

3/29

5/29

4/29

−3/29

= A⁻¹

Step 4 — x = A⁻¹b

            x = (3/29)×7 + (5/29)×19
= 21/29 + 95/29
= 4


            y = (4/29)×7 + (−3/29)×19
= 28/29 − 57/29
= −1
          

✅ x = 4 y = −1
Check eq1: 3(4)+5(−1) = 12−5 = 7 ✓ Check eq2: 4(4)−3(−1) = 16+3 = 19 ✓

🎛️Visual Step-by-Step SolverInteractive

Red = x-coefficients | Yellow = y-coefficients | Green = constants. Steps show each transformation of A into A⁻¹.

📘 Equation 1: ax + by = e

x + y =

📗 Equation 2: cx + dy = f

x + y =

–

Ready

Enter your equations then press ▶ Show Step-by-Step

Orange = diagonal swapped. Purple = off-diagonal negated. Green = final A⁻¹. Watch A transform at each step.

👆Press ▶ Show Step-by-Step to see each determinant built live · Try 3×3 mode to see Dx, Dy, Dz all computed · Change any coefficient and press ▶ again to re-run instantly.

📊

Gaussian Elimination

Use row operations on the augmented matrix [A|b] to reach upper-triangular form, then back-substitute from bottom to top.

🎨 Colour Key — what each colour means in Gaussian Elimination

🔴 Red

Pivot row
Used to eliminate below

🟣 Purple

Row being eliminated
(changing this step)

🔵 Blue

New zero created
(elimination succeeded)

🟢 Green

Constants column (b)
right-hand side

🟡 Gold

Back-substitution
answer cells

💡 The multiplier m = (pivot row element) ÷ (pivot). Subtract m × pivot row from the row being eliminated to create a zero.

📋The Augmented Matrix [A|b]Setup

Write the coefficient matrix A and the constants b side by side, separated by a bar. Row operations on the whole augmented matrix mirror what you'd do algebraically to the equations.

From equations…

          a₁x + b₁y = c₁

          a₂x + b₂y = c₂

…to augmented matrix

a₁

b₁

a₂

b₂

c₁

c₂

⚙️The Three Row OperationsRules

Rᵢ ↔ Rⱼ

🔃

Swap two rows

Used for partial pivoting — put the largest pivot at the top

kRᵢ → Rᵢ

✖️

Scale a row by k ≠ 0

Multiply every element in the row by the same number

Rᵢ + kRⱼ → Rᵢ

➕

Add multiple of Rⱼ to Rᵢ

The key operation — creates zeros below pivots

🎯 Goal: get zeros below every pivot to create upper-triangular form, then back-substitute from the bottom row upwards.

✏️Worked Example — 2×2Full walkthrough

Solve using Gaussian Elimination

Step 1 — Augmented [A|b]

3

5

4

−3

7

19

Step 2 — R₂ → R₂ − (4/3)R₁

m = 4÷3 = 1.33
Pivot row (R₁) = red
Eliminated row (R₂) = purple

3

5

0

−29/3

7

29/3

🔵 Blue = new zero created!

Step 3 — Back-Substitute

            From R₂: (−29/3)y = 29/3

            y = −1

            From R₁: 3x + 5(−1) = 7

            3x = 12

            x = 4

✅ x = 4 y = −1
Check eq1: 3(4)+5(−1) = 12−5 = 7 ✓ Check eq2: 4(4)−3(−1) = 16+3 = 19 ✓

🎛️Visual Step-by-Step SolverInteractive

📘 Equation 1

x + y =

📗 Equation 2

x + y =

–

Ready

Enter your equations then press ▶ Show Step-by-Step

🔴 Red = pivot row | 🟣 Purple = row being eliminated | 🔵 Blue = new zero | 🟡 Gold = back-substitution answer

👆Step through Cofactors → Adjugate → A⁻¹ one matrix at a time · Try 2×2 mode first to see the pattern clearly · Enter a singular system like x+2y=3, 2x+4y=6 — det=0 so no inverse exists.

⚖️

Comparing the Four Methods

Know when to use each method — exam guidance and practical trade-offs.

📊Side-by-Side SummaryReference

Method	How it works	Best when…	Watch out for	Scales to 4×4+?
📐 Determinant	Calculate det(A) = ad−bc (or cofactor expansion)	Checking if solution exists; foundation for other methods	3×3 arithmetic errors	✅ Yes (laborious)
🧮 Cramer's Rule	Replace each column with b; divide det ratios	Exam asks "use determinants"; finding only one unknown	Wrong column replaced; tedious for 3×3	❌ Impractical
🔄 Inverse Method	Find A⁻¹, multiply x = A⁻¹b	Reusing same A with different b; "use matrices"	3×3 cofactor sign errors	❌ Impractical
📊 Gaussian	Row ops on [A\|b] → upper-triangular → back-sub	Any system; most reliable; no method specified	Fraction arithmetic compounds; partial pivot needed	✅ Best choice

🎓Exam GuidanceImportant

"Use Cramer's Rule" → must use determinant method
"Use matrices" → inverse method is expected
No method specified → Gaussian is safest
Always state det(A) first — marks often awarded
Always check by substituting back into originals

🔁Same Problem — All MethodsComparison

System: 3x + 5y = 7, 4x − 3y = 19

det(A): 3×(−3) − 5×4 = −29 Cramer's: D=−29, Dx=−116, Dy=29 → x=4, y=−1 Inverse: A⁻¹=(1/−29)[−3 −5;−4 3] → x=4, y=−1 Gaussian: R₂−(4/3)R₁ → y=−1 → x=4

✅ All methods give the same answer!

👆Press ▶ Start Elimination to watch row ops one at a time · Switch to 3×3 to see two elimination passes · Try x−y=2, 2x+y=7 — follow the pivot row then back-substitute from the bottom.

🎮

Quick Fire Quiz

10 questions across all four methods. Beat your streak!

Question 1/10

Score 0

Streak 🔥 0

✏️

Practice Problems

10 problems across all four methods. Use the hints if stuck, then reveal the full worked solution.

🎯Filter by Method