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x = Dₓ ÷ D · y = D_y ÷ D

Simultaneous equations calculator

Two equations in x and y, solved by Cramer’s rule, with parallel and identical lines told apart rather than lumped together.

First equation: a₁x + b₁y = c₁

Second equation: a₂x + b₂y = c₂

Short answer: x + y = 10 with x − y = 2 gives x = 6 and y = 4. The determinant is 1 × (−1) − 1 × 1 = −2, and dividing the two Cramer determinants by it gives the pair.

Why the determinant matters

The determinant is the pair of lines’ slopes compared: zero means they rise at the same rate, so they never meet or they coincide. Printing D first is what tells you whether a solution exists at all, before any division happens.

Worked example

Four pens and two notebooks cost ₹48; two pens and three notebooks cost ₹34. With x for a pen: 4x + 2y = 48 and 2x + 3y = 34 give x = ₹11 and y = ₹2, and substituting both back gives 44 + 4 = 48 and 22 + 6 = 34.

Simultaneous equation questions

How are the two unknowns found?

Cramer’s rule: D = a₁b₂ − a₂b₁, then x = Dₓ ÷ D and y = D_y ÷ D. When D is non-zero the lines cross once and there is exactly one pair (x, y).

What if the determinant is zero?

Then the lines are parallel or identical, and the two cases are reported separately. A zero determinant with proportional equations means infinitely many solutions; without proportionality there are none. Saying "no solution" for both would be wrong.

How do I check the answer?

Put the pair back into both equations — the page is built so the numbers are easy to substitute, and x = 6, y = 4 in x + y = 10 and x − y = 2 satisfies both. Any solution that fails one equation is a slip in reading the coefficients.

Does it handle three unknowns?

No. A system in three unknowns has three equations and needs a different method; this page covers the 2×2 case, which is the one that appears in school work and in two-variable costing problems.

Related tools

Linear equation calculatorOne unknown, with the steps.Matrix calculator3×3 determinants and inverses.Quadratic equation calculatorWhen one of the equations is squared.