Math

System of Equations Calculator

Solve a two-variable linear system with exact fractions and determinant steps, including parallel and dependent equations.

TRY A WORKED EXAMPLE

Choose a starting point, adjust the values, then calculate.

Use whole-number coefficients from −10,000 to 10,000. The equation format is shown in the input labels.
01

Your inputs

02

Your result

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Using the System of Equations Calculator

Use two equations to find two unknowns, such as item prices or the intersection of straight lines. Write each equation in ax + by = c form first. The second row uses d, e and f to avoid confusing the coefficients.

Worked example

Use the example inputs below to reproduce this result. These are the form's starting values, not recommended targets. Changing your inputs updates your result above; this worked example stays fixed for comparison. Results are rounded for display.

Example inputs

a in ax + by = c
2
b in ax + by = c
1
c in ax + by = c
7
d in dx + ey = f
1
e in dx + ey = f
-1
f in dx + ey = f
2

Example result

Solution
x = 3, y = 1
x
3
y
1
Verification
Exact cross-products satisfy both original equations.
Steps
Start with 2x + (1)y = 7 and 1x + (-1)y = 2. Compute D = a·e − b·d = -3. Compute Dx = c·e − b·f = -9; Dy = a·f − c·d = -3. Divide: x = Dx/D = 3; y = Dy/D = 1.

Method and assumptions

Use 2×2 determinants (Cramer’s rule). If the determinant is zero, compare augmented constraints, including zero rows, to distinguish infinitely many solutions from no solution. Two equations and integer coefficients only.

Reference: openstax.org — method and further reading

Check the answer in the original problem

After solving, substitute the result into the original equation or relationship where possible. This can catch a sign error or a coefficient entered in the wrong position. Keep the exact inputs available when comparing with another tool, since changing a coefficient or rounding an intermediate value can change the problem being solved.

Common question

What does a zero determinant mean?

There is no unique intersection. Compatible dependent equations have infinitely many solutions, while contradictory constraints have none. Zero rows are checked separately.