Use signed whole numbers of up to 300 digits. Commas, spaces, or underscores may group digits in sets of three.
For the starting equation 15x + 21y = 84, select Solve equation to get a particular solution and the complete family.
Let d = gcd(a, b), using a non-negative greatest common divisor. When a and b are not both zero, the equation has an integer solution exactly when d divides c.
ax + by = c is solvable ⇔ gcd(a,b) | c
If d does not divide c, no pair of integers (x, y) can satisfy the equation.
The reason is Bézout’s identity. The extended Euclidean algorithm finds integers s and t with as + bt = d. If c = dq, multiplying by q gives the particular solution x0 = sq, y0 = tq.
Suppose (x0, y0) is one solution and d = gcd(a, b). Every integer solution is:
x = x0 + (b/d)k
y = y0 − (a/d)k
k ∈ ℤ
Changing k adds b/d to x and subtracts a/d from y. The equation stays balanced because the change in its left side is:
a(b/d)k + b(−a/d)k = 0
The calculator chooses a compact starting pair by shifting the extended-Euclidean result so x0 is as close to zero as possible when b ≠ 0.
gcd(15, 21) = 3, and 3 divides 84 because 84 ÷ 3 = 28. Integer solutions therefore exist.
Bézout’s identity gives 15(3) + 21(−2) = 3. Multiplying by 28 first gives (84, −56), then shifting within the family gives the smaller particular solution (0, 4).
x = 0 + 7k
y = 4 − 5k
For k = 2, (x, y) = (14, −6):
15(14) + 21(−6)
= 210 − 126
= 84
| Equation form | Integer-solution rule |
|---|---|
| a and b both nonzero | Solutions exist exactly when gcd(a, b) divides c; the general two-variable formula applies. |
| a = 0, b ≠ 0 | y = c/b when b divides c, while x may be any integer. |
| a ≠ 0, b = 0 | x = c/a when a divides c, while y may be any integer. |
| a = b = 0, c = 0 | Every integer pair (x, y) is a solution. |
| a = b = 0, c ≠ 0 | No solution, because the left side is always zero. |
When a and b are not both zero, solutions exist exactly when gcd(a, b) divides c. The solver checks this before constructing a solution.
A solvable equation with at least one nonzero coefficient has infinitely many integer solutions. The parameter k runs through all integers to generate them.
It chooses one member of the solution family. Replacing k with any other integer gives another exact solution without changing the equation.
Yes. All four inputs accept signed integers. The result uses the original signs and separately explains the all-zero coefficient case.
Particular solutions are not unique. Two correct starting pairs differ by a shift of (b/d, −a/d) times an integer. This solver selects a compact representative.
Each input may contain up to 300 digits after grouping separators are removed. JavaScript BigInt arithmetic keeps every displayed integer exact; long Euclidean tables are abbreviated on screen.
Yes. Validation and arithmetic happen entirely in your browser. The tool does not upload, store, or add your coefficients to the URL.
