Solving Simultaneous Linear Equations

Solves a₁x + b₁y = c₁ and a₂x + b₂y = c₂ for x and y. When a₁b₂ − a₂b₁ is zero there is no unique solution.

Explanation

Two linear equations in two unknowns are solved together for the pair $(x, y)$ that satisfies both at once.

a_1 x + b_1 y = c_1, \quad a_2 x + b_2 y = c_2

Substitution and elimination both work, but Cramer's rule writes the answer straight from the coefficients.

x = \dfrac{c_1 b_2 - c_2 b_1}{a_1 b_2 - a_2 b_1}, \quad y = \dfrac{a_1 c_2 - a_2 c_1}{a_1 b_2 - a_2 b_1}

$a_1$ , $b_1$ , $c_1$ — the coefficients and the right-hand side of the first equation
$a_2$ , $b_2$ , $c_2$ — the same for the second equation
the shared denominator $a_1 b_2 - a_2 b_1$ is the determinant of the coefficient matrix

Example

Solve $2x + 3y = 12$ together with $x - y = 1$ , so $a_1 = 2$ , $b_1 = 3$ , $c_1 = 12$ , $a_2 = 1$ , $b_2 = -1$ , $c_2 = 1$ . The denominator first:

a_1 b_2 - a_2 b_1 = 2 \times (-1) - 1 \times 3 = -5

x = \dfrac{12 \times (-1) - 1 \times 3}{-5} = \dfrac{-15}{-5} = 3, \quad y = \dfrac{2 \times 1 - 1 \times 12}{-5} = \dfrac{-10}{-5} = 2

So $x = 3$ and $y = 2$ . Substituting back, $2 \times 3 + 3 \times 2 = 12$ and $3 - 2 = 1$ : both equations hold.

When the denominator $a_1 b_2 - a_2 b_1$ is zero there is no unique solution, and the calculator reports it.
Geometrically the two lines are then either parallel, never meeting, so there is no solution at all, or identical, so every point on the line is a solution. Either way the coefficient ratios match.
Rearrange each equation into the form $ax + by = c$ before reading off coefficients. Something like $y = 2x + 1$ has to be rewritten as $-2x + y = 1$ .
Always substitute your answer back into both original equations as a check.