Matemática

Cramer's Rule Calculator: Solve Any 2×2 Linear System

Solve ax+by=e, cx+dy=f instantly with Cramer's Rule. Enter the 6 coefficients and get x, y, and the determinant — plus a step-by-step explanation.

🗓️ Updated June 2026 Reviewed by Martín Rodríguez

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Reviewed by: Martín Rodríguez (editorial policy ) · Last reviewed: Jun 22, 2026

a (coeff. of x, eq. 1)*

b (coeff. of y, eq. 1)*

e (right-hand side, eq. 1)*

c (coeff. of x, eq. 2)*

d (coeff. of y, eq. 2)*

f (right-hand side, eq. 2)*

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Cramer's Rule is a closed-form method for solving 2×2 linear systems using determinants. Given the system ax + by = e and cx + dy = f, the solution is x = (ed − bf)/(ad − bc) and y = (af − ce)/(ad − bc), where D = ad − bc is the determinant of the coefficient matrix. If D ≠ 0, the system has exactly one unique solution (the two lines intersect at one point). If D = 0, the system is either inconsistent (parallel lines, no solution) or dependent (same line, infinitely many solutions). Enter your six coefficients below to solve instantly.

When to use this calculator

Finding the exact intersection point of two straight lines in coordinate geometry (e.g., supply and demand curves in economics)
Solving electrical circuit equations from Kirchhoff's Voltage Law where two loop currents are unknown
Computing 2D affine transformation coefficients (scale/shear) when two control-point pairs are given
Determining break-even quantities for two competing products with shared resource constraints in operations research

Cramer's Rule answer key — solved 2×2 systems with their determinants

Plug these coefficients into the calculator to verify it, or check your hand-worked solution against a known-correct result. Every D, x and y below is exact.

System	a	b	c	d	e	f	D = ad − bc	x	y
x + y = 5 ; 2x + 3y = 13	1	1	2	3	5	13	1	2	3
2x + 3y = 12 ; 4x − y = 10	2	3	4	-1	12	10	-14	3	2
5x + 2y = 16 ; 3x + 7y = 27	5	2	3	7	16	27	29	2	3
x − 4y = −9 ; 6x + y = 4	1	-4	6	1	-9	4	25	0.28	2.32
4I₁ − 2I₂ = 12 ; −2I₁ + 6I₂ = 0	4	-2	-2	6	12	0	20	3.6	1.2
x + y = 3 ; 2x + 2y = 8	1	1	2	2	3	8	0	—	no solution (parallel)
x + y = 3 ; 2x + 2y = 6	1	1	2	2	3	6	0	—	∞ solutions (same line)

Solutions computed as x = (ed − bf)/D and y = (af − ec)/D with D = ad − bc. When D = 0 the system is degenerate: if Dₓ or D_y is non-zero the lines are parallel (no solution); if both are zero the lines coincide (infinitely many solutions). The Kirchhoff row (I₁, I₂) shows two mesh currents in amps from a resistor network.

How it works

How Cramer's Rule Works

Given the 2×2 linear system:

a·x + b·y = e
c·x + d·y = f

Step 1 — Compute the coefficient determinant (D):

D = |a  b| = a·d − b·c
    |c  d|

Step 2 — Compute Dₓ (replace the x-column with the constants):

Dₓ = |e  b| = e·d − b·f
     |f  d|

Step 3 — Compute D_y (replace the y-column with the constants):

D_y = |a  e| = a·f − e·c
      |c  f|

Step 4 — Apply Cramer's Rule:

x = Dₓ / D = (e·d − b·f) / (a·d − b·c)
y = D_y / D = (a·f − e·c) / (a·d − b·c)

If D = 0 and Dₓ ≠ 0 or D_y ≠ 0 → inconsistent (parallel lines, no solution).
If D = 0 and Dₓ = 0 and D_y = 0 → dependent (same line, infinitely many solutions).

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Quick Reference Table — Common Systems

System	a	b	c	d	e	f	D	x	y
x+y=5, 2x+3y=13	1	1	2	3	5	13	1	2	3
3x−y=4, x+2y=13	3	−1	1	2	4	13	7	3	5
4I₁−2I₂=12, −2I₁+6I₂=0	4	−2	−2	6	12	0	20	3.6	1.2
0.5x+1.5y=4, 2x−y=1	0.5	1.5	2	−1	4	1	−3.5	≈1.571	≈2.143
x+y=3, 2x+2y=8	1	1	2	2	3	8	0	—	— (no solution)
x+y=3, 2x+2y=6	1	1	2	2	3	6	0	—	— (∞ solutions)

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Determinant Interpretation

D value	Dₓ / D_y	Geometric meaning	Solution type
D ≠ 0	Any	Lines intersect at exactly one point	Unique solution
D = 0	Dₓ = 0, D_y = 0	Lines are identical (coincident)	Infinite solutions
D = 0	Dₓ ≠ 0 or D_y ≠ 0	Lines are parallel (never meet)	No solution

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Detailed Worked Examples

Example 1 — Classic textbook system

System: x + y = 5 and 2x + 3y = 13

D = (1)(3) − (1)(2) = 3 − 2 = 1

Dₓ = (5)(3) − (1)(13) = 15 − 13 = 2 → x = 2/1 = 2

D_y = (1)(13) − (5)(2) = 13 − 10 = 3 → y = 3/1 = 3

Solution: x = 2, y = 3 ✓ (verify: 2+3=5 ✓, 4+9=13 ✓)

Example 2 — Kirchhoff circuit loop equations

System: 4I₁ − 2I₂ = 12 and −2I₁ + 6I₂ = 0

D = (4)(6) − (−2)(−2) = 24 − 4 = 20

Dₓ = (12)(6) − (−2)(0) = 72 − 0 = 72 → I₁ = 72/20 = 3.6 A

D_y = (4)(0) − (12)(−2) = 0 + 24 = 24 → I₂ = 24/20 = 1.2 A

Example 3 — No solution (parallel lines)

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

D = (2)(2) − (4)(1) = 4 − 4 = 0

Dₓ = (6)(2) − (4)(5) = 12 − 20 = −8 ≠ 0

Result: Inconsistent — no solution (lines are parallel, different y-intercepts)

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Common Mistakes

1. Swapping rows vs. columns when forming Dₓ and D_y. The constants (e, f) replace the column of the target variable, not a row.
2. Sign errors. Remember: D = ad − bc, NOT ad + bc. A single missed minus sign flips the entire solution.
3. Dividing by zero without checking D first. Always verify D ≠ 0 before computing x and y.
4. Assuming D = 0 always means no solution. Check Dₓ and D_y to distinguish inconsistent from dependent systems.
5. Not rewriting in standard form. If the system is written as y + 2x = 7, rewrite it as 2x + y = 7 before extracting coefficients.

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Worked Example

System: x + y = 5 and 2x + 3y = 13

Enter: a=1, b=1, e=5, c=2, d=3, f=13

D = (1)(3) − (1)(2) = 3 − 2 = 1

x = (5·3 − 1·13) / 1 = (15 − 13) / 1 = 2

y = (1·13 − 5·2) / 1 = (13 − 10) / 1 = 3

Verify: 2+3=5 ✓ and 2(2)+3(3)=13 ✓

x = 2, y = 3

Frequently asked questions

What is Cramer's Rule and how do you apply it to a 2×2 system?

Cramer's Rule (published by Gabriel Cramer in 1750) expresses each variable as a ratio of two determinants. For ax+by=e, cx+dy=f: first compute D = ad−bc. Then x = (ed−bf)/D and y = (af−ec)/D. If D ≠ 0 you get exactly one solution; if D = 0 the system is degenerate.

What is the formula for solving a 2×2 system with Cramer's Rule?

Given ax+by=e and cx+dy=f, the formulas are: D = ad−bc, x = (ed−bf)/D, y = (af−ec)/D. These are derived by replacing the target variable's column in the coefficient matrix with the constant column (e, f) and dividing by D.

When does a 2×2 system have no solution vs. infinitely many solutions?

Both cases arise when D = ad−bc = 0. If D = 0 and at least one of Dₓ=(ed−bf) or D_y=(af−ec) is non-zero, the system is inconsistent (parallel lines, no solution). If D = 0 and both Dₓ = 0 and D_y = 0, the system is dependent (same line, infinitely many solutions). Always check both auxiliary determinants.

How do I verify the solution x, y is correct?

Substitute x and y back into both original equations. For x+y=5, 2x+3y=13 with solution x=2, y=3: check 2+3=5 ✓ and 2(2)+3(3)=4+9=13 ✓. Both equations must balance — one equation passing is not sufficient.

Can Cramer's Rule handle decimal or fractional coefficients?

Yes. The rule works for any real-number coefficients. For example, 0.5x+1.5y=4 and 2x−y=1 gives D=(0.5)(−1)−(1.5)(2)=−3.5, x=(4·(−1)−1.5·1)/(−3.5)=−5.5/−3.5≈1.571, y=(0.5·1−4·2)/(−3.5)=−7.5/−3.5≈2.143. The calculator handles all real-valued inputs automatically.

Is Cramer's Rule the most efficient method for 2×2 systems?

For exactly 2×2 systems, Cramer's Rule, substitution, and elimination all require roughly the same arithmetic effort. However, for n×n systems with n ≥ 4, Gaussian elimination (O(n³)) drastically outperforms Cramer's Rule (O(n·n!)), which is why numerical software like MATLAB or NumPy uses LU decomposition instead. Cramer's Rule excels when you need a clean, symbolic closed-form answer for one variable at a time.

What does the determinant D represent geometrically?

For a 2×2 matrix with row vectors u=(a,b) and v=(c,d), |D| = |ad−bc| equals the area of the parallelogram spanned by those two vectors. When D=0 the vectors are collinear (linearly dependent), meaning the lines either overlap or are parallel — matching the degenerate cases. This connects Cramer's Rule to cross products and area calculations in physics and engineering.

How do I enter a system where a variable has an implied coefficient of 1 or is missing?

An implied coefficient of 1 is still the number 1. For example, x + 3y = 7 is entered as a=1, b=3, e=7. If a variable is missing (e.g., −y = 2, meaning 0·x + (−1)·y = 2), enter a=0, b=−1, e=2. Always make every coefficient explicit before inputting values. The system must be in the form ax+by=e before extracting the six coefficients.

Why does the calculator show determinant 0 for x+y=3, 2x+2y=6?

Because D = (1)(2)−(1)(2) = 2−2 = 0. The second equation is exactly 2 times the first (multiply x+y=3 by 2 to get 2x+2y=6), so both equations describe the same line. Since Dₓ=(3·2−1·6)=0 and D_y=(1·6−3·2)=0, the system is dependent with infinitely many solutions y=(3−x)/2 for any real x.

Sources & references

Methodology & trust

Editorial

Calculadora de matemática revisada por el equipo editorial de Hacé Cuentas, contrastada con Wikipedia — Cramer's Rule, según nuestra política editorial y metodología.

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Última revisión: June 22, 2026. Los parámetros se verifican periódicamente con las fuentes citadas.

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📌 How to cite this calculator

Rodríguez, M. (2026). Cramer's Rule Calculator: Solve Any 2×2 Linear System. Hacé Cuentas. https://hacecuentas.com/2x2-linear-system-cramer-rule

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