The setup
A system of two linear equations in two unknowns:
a₁x + b₁y = c₁ and a₂x + b₂y = c₂
The solution is the point where the two lines intersect, and the determinant D = a₁b₂ − a₂b₁ decides which of three things happens before you solve anything.
The three cases, decided by the determinant
- D ≠ 0 → one unique solution. The lines cross at a single point.
- D = 0 and the ratios a₁/a₂ = b₁/b₂ = c₁/c₂ → infinitely many solutions. The equations describe the same line.
- D = 0 but the ratios differ → no solution. The lines are parallel and distinct.
The geometric reading is in the linear lines guide: the slopes are a₁/b₁ and a₂/b₂, and D = 0 is exactly the statement that they are equal.
Worked example: unique solution
2x + 3y = 7 and x − y = 1.
D = (2)(−1) − (1)(3) = −2 − 3 = −5, so a unique solution exists.
x = (c₁b₂ − c₂b₁) ÷ D = (7 × −1 − 1 × 3) ÷ −5 = (−7 − 3) ÷ −5 = 2
y = (a₁c₂ − a₂c₁) ÷ D = (2 × 1 − 1 × 7) ÷ −5 = (2 − 7) ÷ −5 = 1
Check by substitution: 2(2) + 3(1) = 7 ✓ and 2 − 1 = 1 ✓. The calculator returns both checks automatically, which is worth doing by hand once — a single transcribed sign error produces a plausible-looking wrong answer, and the check catches it immediately.
Worked example: infinitely many solutions
2x + 4y = 6 and x + 2y = 3.
D = (2)(2) − (1)(4) = 4 − 4 = 0. Ratios: 2/1 = 2, 4/2 = 2, 6/3 = 2 — all equal, so the second equation is just the first halved. Every point on that line is a solution.
Worked example: no solution
2x + 4y = 6 and x + 2y = 5.
D = 0 again, but the ratios are 2/1 = 2, 4/2 = 2, and 6/5 = 1.2. Since the coefficients match while the constants do not, the lines are parallel and never meet. This is the case students most often misread as "infinitely many", so it is worth checking the constants explicitly whenever D = 0.
Elimination by hand
The calculator uses Cramer's rule, but the method taught in class is elimination, and it is worth being able to do it manually:
2x + 3y = 7 and x − y = 1. Multiply the second equation by 2 to match the x coefficients: 2x − 2y = 2. Subtract: (2x + 3y) − (2x − 2y) = 7 − 2, giving 5y = 5, so y = 1. Substitute back: 2x + 3 = 7, so x = 2.
Two details cause most errors. The subtraction step must be consistent throughout — once you subtract y terms you must keep doing so, not switch to adding. And when the coefficients do not match, multiply one whole equation by a constant, never individual terms.
Three-variable systems
A third equation and variable extends the method: eliminate down from three equations to two, then to one. The structure is identical and the arithmetic grows quickly, which is why in practice 3×3 systems are solved with matrices or software. The relevant tool is the determinant: a 3×3 system has a unique solution exactly when its determinant is non-zero.
Frequently asked questions
How do I solve a system of two equations?
Use elimination: match a variable's coefficient by multiplying one equation, subtract to eliminate it, then solve the remaining single variable and substitute back. Cramer's rule gives the same answer in one step: x = (c₁b₂ − c₂b₁) ÷ D with D = a₁b₂ − a₂b₁.
What does a determinant of zero mean?
It means the system has either no solution or infinitely many, never exactly one. If the coefficient ratios all match, the equations are the same line (infinitely many solutions); if only the coefficients match but the constants do not, the lines are parallel (no solution).
How do I know if the lines are parallel or identical?
When the determinant is zero, compare the ratios of the coefficients. Equal a, b and c ratios mean the equations describe the same line; equal a and b ratios but different c ratios mean parallel distinct lines with no intersection.
Can a system have more than one solution?
For two linear equations in two unknowns, the possibilities are exactly one, zero, or infinitely many — never a finite number greater than one. That is a consequence of two straight lines meeting at most once. For non-linear equations, more exotic possibilities exist, which is why the linear case is worth understanding first.