Sum of two numbers if the original ratio and new ratio obtained by adding a given number to each number is given

Given a ratio a : b of two unknown numbers. When both the numbers are incremented by a given integer x, the ratio becomes c : d. The task is to find the sum of the two numbers.
Examples:

Input: a = 2, b = 3, c = 8, d = 9, x = 6 
Output: 5 
Original numbers are 2 and 3 
Original ratio = 2:3 
After adding 6, ratio becomes 8:9 
2 + 3 = 5
Input: a = 1, b = 2, c = 9, d = 13, x = 5 
Output: 12 

Approach: Let the sum of the numbers be S. Then, the numbers can be (a * S)/(a + b) and (b * S)/(a + b). 
Now, as given: 

(((a * S) / (a + b)) + x) / (((b * S) / (a + b)) + x) = c / d
or ((a * S) + x * (a + b)) / ((b * S) + x * (a + b)) = c / d
So, S = (x * (a + b) * (c - d)) / (ad - bc)

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the sum of numbers
// which are in the ration a:b and after
// adding x to both the numbers
// the new ratio becomes c:d
double sum(double a, double b, double c, double d, double x)
{
    double ans = (x * (a + b) * (c - d)) / ((a * d) - (b * c));
    return ans;
}

// Driver code
int main()
{
    double a = 1, b = 2, c = 9, d = 13, x = 5;

    cout << sum(a, b, c, d, x);

    return 0;
}

Java

// Java implementation of the above approach
class GFG
{

    // Function to return the sum of numbers
    // which are in the ration a:b and after
    // adding x to both the numbers
    // the new ratio becomes c:d
    static double sum(double a, double b,
                      double c, double d,
                      double x)
    {
        double ans = (x * (a + b) * (c - d)) /
                         ((a * d) - (b * c));
        return ans;
    }

    // Driver code
    public static void main (String[] args)
    {
        double a = 1, b = 2, c = 9, d = 13, x = 5;

        System.out.println(sum(a, b, c, d, x));
    }
}

// This code is contributed by ihritik

Python3

# Python3 implementation of the approach

# Function to return the sum of numbers
# which are in the ration a:b and after
# adding x to both the numbers
# the new ratio becomes c:d
def sum(a, b, c, d, x):

    ans = ((x * (a + b) * (c - d)) /
               ((a * d) - (b * c)));
    return ans;

# Driver code
if __name__ == '__main__':

    a, b, c, d, x = 1, 2, 9, 13, 5;

    print(sum(a, b, c, d, x));

# This code is contributed by PrinciRaj1992

C#

// C# implementation of the above approach
using System;

class GFG
{

    // Function to return the sum of numbers
    // which are in the ration a:b and after
    // adding x to both the numbers
    // the new ratio becomes c:d
    static double sum(double a, double b,
                      double c, double d,
                      double x)
    {
        double ans = (x * (a + b) * (c - d)) /
                         ((a * d) - (b * c));
        return ans;
    }

    // Driver code
    public static void Main ()
    {
        double a = 1, b = 2,
               c = 9, d = 13, x = 5;

        Console.WriteLine(sum(a, b, c, d, x));
    }
}

// This code is contributed by ihritik

Javascript

<script>

// javascript implementation of the above approach

// Function to return the sum of numbers
// which are in the ration a:b and after
// adding x to both the numbers
// the new ratio becomes c:d
function sum(a , b, c , d, x)
{
    var ans = (x * (a + b) * (c - d)) /
                     ((a * d) - (b * c));
    return ans;
}

// Driver code

var a = 1, b = 2, c = 9, d = 13, x = 5;

document.write(sum(a, b, c, d, x));


// This code is contributed by 29AjayKumar

</script>

Output:

12

Time Complexity: O(1)

Auxiliary Space: O(1)