# Find smallest values of x and y such that ax – by = 0

Given two values ‘a’ and ‘b’ that represent coefficients in “ax – by = 0”, find the smallest values of x and y that satisfy the equation. It may also be assumed that x > 0, y > 0, a > 0 and b > 0.

```Input: a = 25, b = 35
Output: x = 7, y = 5
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A Simple Solution is to try every possible value of x and y starting from 1, 1 and stop when the equation is satisfied.

A Direct Solution is to use Least Common Multiple (LCM). LCM of ‘a’ and ‘b’ represents the smallest value that can make both sides equal. We can find LCM using below formula.

`   LCM(a, b) = (a * b) / GCD(a, b) `

Greatest Common Divisor (GCD) can be computed using Euclid’s algorithm.

## C++

 `// C++ program to find the smallest values of x and y that ` `// satisfy "ax - by = 0" ` `#include ` `using` `namespace` `std; ` ` `  `// To find GCD using Eculcid's algorithm ` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(b == 0) ` `        ``return` `a; ` `    ``return` `(gcd(b, a % b)); ` `} ` ` `  `// Prints smallest values of x and y that ` `// satisfy "ax - by = 0" ` `void` `findSmallest(``int` `a, ``int` `b) ` `{ ` `    ``// Find LCM ` `    ``int` `lcm = (a * b) / gcd(a, b); ` ` `  `    ``cout << ``"x = "` `<< lcm / a ` `         ``<< ``"\ny = "` `<< lcm / b; ` `} ` ` `  `// Driver program ` `int` `main() ` `{ ` `    ``int` `a = 25, b = 35; ` `    ``findSmallest(a, b); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find the smallest values of ` `// x and y that satisfy "ax - by = 0" ` `class` `GFG { ` ` `  `    ``// To find GCD using Eculcid's algorithm ` `    ``static` `int` `gcd(``int` `a, ``int` `b) ` `    ``{ ` ` `  `        ``if` `(b == ``0``) ` `            ``return` `a; ` `        ``return` `(gcd(b, a % b)); ` `    ``} ` ` `  `    ``// Prints smallest values of x and y that ` `    ``// satisfy "ax - by = 0" ` `    ``static` `void` `findSmallest(``int` `a, ``int` `b) ` `    ``{ ` ` `  `        ``// Find LCM ` `        ``int` `lcm = (a * b) / gcd(a, b); ` ` `  `        ``System.out.print(``"x = "` `+ lcm / a ` `                         ``+ ``"\ny = "` `+ lcm / b); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `a = ``25``, b = ``35``; ` `        ``findSmallest(a, b); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# Python program to find the ` `# smallest values of x and y that ` `# satisfy "ax - by = 0" ` ` `  `# To find GCD using Eculcid's algorithm ` `def` `gcd(a, b): ` `    ``if` `(b ``=``=` `0``): ` `        ``return` `a ` `    ``return``(gcd(b, a ``%` `b)) ` ` `  `# Prints smallest values of x and y that ` `# satisfy "ax - by = 0" ` `def` `findSmallest(a, b): ` ` `  `    ``# Find LCM ` `    ``lcm ``=` `(a ``*` `b)``/``gcd(a, b) ` `    ``print``(``"x ="``, lcm ``/` `a, ``"\ny = "``, lcm ``/` `b) ` ` `  `# Driver code ` `a ``=` `25` `b ``=` `35` `findSmallest(a, b) ` ` `  `# This code is contributed ` `# by Anant Agarwal. `

## C#

 `// C# program to find the smallest ` `// values of x and y that ` `// satisfy "ax - by = 0" ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// To find GCD using ` `    ``// Eculcid's algorithm ` `    ``static` `int` `gcd(``int` `a, ``int` `b) ` `    ``{ ` ` `  `        ``if` `(b == 0) ` `            ``return` `a; ` `        ``return` `(gcd(b, a % b)); ` `    ``} ` ` `  `    ``// Prints smallest values of x and ` `    ``// y that satisfy "ax - by = 0" ` `    ``static` `void` `findSmallest(``int` `a, ``int` `b) ` `    ``{ ` ` `  `        ``// Find LCM ` `        ``int` `lcm = (a * b) / gcd(a, b); ` ` `  `        ``Console.Write(``"x = "` `+ lcm / a + ``"\ny = "` `+ lcm / b); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `a = 25, b = 35; ` `        ``findSmallest(a, b); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007. `

## PHP

 ` `

Output:

```x = 7
y = 5
```

The above code for findSmallest() can be reduced:

```Since ax - by = 0,
ax = by, which means x/y = b/a
So we can calculate gcd and directly do as -

Value of x = b / gcd;
Value of y = a / gcd; ```

 `// Prints smallest values of x and y that ` `// satisfy "ax - by = 0" ` `void` `findSmallest(``int` `a, ``int` `b) ` `{ ` `    ``// Find GCD ` `    ``int` `g = gcd(a, b); ` ` `  `    ``cout << ``"x = "` `<< b / g ` `         ``<< ``"\ny = "` `<< a / g; ` `} `

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Improved By : Sam007, jit_t

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