Q1. Solve 2x + 3y = 13 and x − y = −1.
From the second: x = y − 1
Substitute: 2(y − 1) + 3y = 13 ⇒ 5y = 15 ⇒ y = 3
x = 3 − 1 = 2. So (x, y) = (2, 3).
Two equations, two unknowns. Enter the coefficients and solve for x and y instantly — then learn every method and the consistency conditions.
Enter a₁x + b₁y = c₁ and a₂x + b₂y = c₂ to solve for x and y.
If a₁b₂ − a₂b₁ = 0 the lines are parallel or the same, so there is no unique solution.
Q1. Solve 2x + 3y = 13 and x − y = −1.
From the second: x = y − 1
Substitute: 2(y − 1) + 3y = 13 ⇒ 5y = 15 ⇒ y = 3
x = 3 − 1 = 2. So (x, y) = (2, 3).
| Quantity | Formula |
|---|---|
| Cross-multiplication (x) | x = (b₁c₂ − b₂c₁)/(a₁b₂ − a₂b₁) |
| Cross-multiplication (y) | y = (c₁a₂ − c₂a₁)/(a₁b₂ − a₂b₁) |
| Unique solution if | a₁/a₂ ≠ b₁/b₂ |
Exactly one (intersecting lines), infinitely many (coincident lines) or none (parallel lines).
When a₁/a₂ ≠ b₁/b₂, i.e. the lines intersect at one point.
Substitution, elimination and cross-multiplication.
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