Thread: Simultaneous linear Equations

I need to make a Console Application which can calculate the values of the Constants " X " and " Y " for the 2 equations ( ax + by = c) and ( dx + ey = f ) .. Of Course the program will ask the user to enter the values for the variables ( a , b , c , d , e , f ). I've tried to make some code, but got stuck at the core of the code. Would anybody please help me?

my Code:

Code:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace _1st_Exercise
{
    class Program
    {
        static void Main(string[] args)
        {
            //Getting the constants
            Console.WriteLine("\"dx + ey = f\" ,, \"ax +by = c\"\n");
            Console.WriteLine("Please enter\n");
            Console.Write("F= ");
            int F;
            F = Convert.ToInt32(Console.ReadLine());
            Console.Write("\nE= ");
            int E;
            E = Convert.ToInt32(Console.ReadLine());
            Console.Write("\nD= ");
            int D;
            D = Convert.ToInt32(Console.ReadLine());
            Console.Write("\nC= ");
            int C;
            C = Convert.ToInt32(Console.ReadLine());
            Console.Write("\nB= ");
            int B;
            B = Convert.ToInt32(Console.ReadLine());
            Console.Write("\nA= ");
            int A;
            A = Convert.ToInt32(Console.ReadLine());

            
            //Code core
            int x;
            x = (A - D);
            
            int y;
            y = (B - E);
            
            
            //Output
            Console.WriteLine("\n\nX = " + x );
            Console.WriteLine("\nY = " + y );
            
            
            
            
            //Exit program
            Console.WriteLine("\n\nPress any key to exit");
            Console.ReadKey();
        
        }
    }
}

You can brute force it in a nested for loop by running through, and checking, each combination of X and Y. Probably not the fastest way to do this though. To give you a rough idea:

Code:

for(y=0; y<100; y++)
    for(x=0; x<100; x++)
        if((A*x + B*y) == C)
            if((D*x + E*y) == F)
                 break;

Would some one Put the Whole code please?

That's my new Code .. is it ok ?

Code:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace _1st_Exercise
{
    class Program
    {
        static void Main(string[] args)
        {
            //Getting the constants
            Console.WriteLine("\"dx + ey = f\" ,, \"ax +by = c\"\n");
            Console.WriteLine("Please enter\n");
            Console.Write("F= ");
            int F;
            F = Convert.ToInt32(Console.ReadLine());
            Console.Write("\nE= ");
            int E;
            E = Convert.ToInt32(Console.ReadLine());
            Console.Write("\nD= ");
            int D;
            D = Convert.ToInt32(Console.ReadLine());
            Console.Write("\nC= ");
            int C;
            C = Convert.ToInt32(Console.ReadLine());
            Console.Write("\nB= ");
            int B;
            B = Convert.ToInt32(Console.ReadLine());
            Console.Write("\nA= ");
            int A;
            A = Convert.ToInt32(Console.ReadLine());

            
            //Code core
            for (int y = 0; y < 100; y++)
            {
                for (int x = 0; x < 100; x++)
                {
                    if ((A * x + B * y) == C)
                    {
                        if ((D * x + E * y) == F)
                        {



                            //Output
                            Console.WriteLine("\n\nX = " + x);
                            Console.WriteLine("\nY = " + y);
                        }
                    }
                }
            }
            
            
            
            
            //Exit program
            Console.WriteLine("\n\nPress any key to exit");
            Console.ReadKey();
        
        }
    }
}

Well, you can test the code if it works, and if it works, it's working. Not so hard?

Originally Posted by Livijn

Well, you can test the code if it works, and if it works, it's working. Not so hard?

haha, that was one fun/good thing you said there. I will use it in the future.