Thread: Writing a program for system of linear equations.

So i was assigned to write a program that would print the answer to the system of two linear equation.
the form of ax+by=c and dx+ey=f
I wrote this program, but it is incorrect since I tried it out on my terminal and I keep getting incorrect values. I would appreciate some help on what I'm doing incorrectly -- my algebra or my programming. Thank You.

import java.util.Scanner;

public class Quest4 {
public static void main(String[] args) {

Scanner input = new Scanner(System.in);

System.out.print("Enter six numbers: ");
double a = input.nextDouble();
double b = input.nextDouble();
double c = input.nextDouble();
double d = input.nextDouble();
double e = input.nextDouble();
double f = input.nextDouble();

if ((a * d) - (b * c) == 0) {
System.out.println("The equation has no solution");
} else {
double x = (e * d - b * f) / (a * d - b * c);
double y = (a * f - e * c) / (a * d - b * c);
System.out.println(" x is " + x + " and " + " y is " + y);
}
}
}

Wow, deja vu right there..... Any way, the simplest way of solving a linear system in the form of y=mx+b is to sub the y value of the first equation into the second, then solve for x. Then using the x value, find y, by subbing it in either equation.

If you had to derive the equations for x and y, can you show or explain how you got what you programmed? If you didn't derive them, where did they come from?

Sorry about that, but I had asked a friend for help on that equation.
I originally had
double x = a * ( c - ( b * ( ( f - d ) / e ) ) );
double y = e * ( f - ( d * ( ( e - b ) / a ) ) );
but that neither worked nor was it even close to what x and y should be.

As you guessed, your equations aren't right. This isn't an Algebra Forum (even though we all love it), but another way to find equations for x and y from the two equations you have is to solve each for x, set the results equal to each other, and then solve for y. If you do that, you should get:

y = (cd - af) / (bd - ae)

Once you have a value for y, you can plug that into either of the original equations to solve for x.

(OR, you could also solve both of the original equations for y, set the results equal to each other, and solve for x.)

How can you tell if there is no solution? If the denominator = 0 in the y equation above, there is no solution, because . . . well, you should know the answer to that.

I recommend you follow either approach I've outlined (or both) and derive the equation I gave above or the similar one for x, because your teacher apparently thinks it's an important part of this class, and you'll probably see something similar again.

You can test my equation to solve for y and then x using the two equations:

x + 2y = 13
3x - 6y = -21

Keep coding!

I cannot thank you enough. I apologize for the problem being an algebraic error the whole time. Thanks!

Hello JavaA123Chris,
Your program is perfect. It will give correct results.
If you believe its giving wrong results provide any instance please.
Note: You are solving the equations using Matrix-Inversion method.

Syed.

Originally Posted by JavaA123Chris

the form of ax+by=c and dx+ey=f
double x = (e * d - b * f) / (a * d - b * c);
double y = (a * f - e * c) / (a * d - b * c);
}

I know this is a little late. But your program works if the equations are of the form
ax+by=e
cx+dy=f

do a little math and we get

ax+by=e <=> ax=e-by <=> x=(e-by)/a
cx+dy=f <=> c((e-by)/a)+dy=f <=> ce-cby+ady=af <=> fa-ce=y(ad-bc) <=>
y=(af-ce)/(ad-bc)
x=(e-by)/a
x=(e-b((af-ce)/(ad-bc))/a <=> x=(e/a)+(b/a)((af-ce)/(ad-bc)) <=>
x=(e/a)((ad-bc)/(ad-bc))+(b/a)((af-ce)/(ad-bc)) <=>
x=(ade-bce)/(a(ad-bc))+(baf-bce)/(a(ad-bc)) <=>
x=(ade-bce+baf-bce)/(a(ad-bc)) <=> x=(ade+baf)/(a(ad-bc)) <=>
x=(de+bf)/(ad-bc)
ad-bc=/=0