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(Algebra: 2 x 2 linear equations) Design a class named LinearEquation for a 2 x

ID: 3577885 • Letter: #

Question

(Algebra: 2 x 2 linear equations) Design a class named LinearEquation for a 2 x 2 system of linear equations: by ar ad bc The class contains: Private data fields a b, c, d, e, and f A constructor with the arguments fora, b, c, d, e, and f Six get functions for a b,c, d, e, and f A function named issolvable() that returns true if ad bc is not 0 Functions getX0 and getY0 that return the solution for the equation. Draw the UML diagram for the class and then implement the class. Write a test program that prompts the user to enter a, b,c, d, e, and fand displays the result. If ad-bc is 0, report that "The equation has no solution. See Programming Exercise for sample runs. (Algebra: solve 2 x 2 linear equations) You can use Cramer's rule to solve the following 2 x 2 system of linear equation: Write a program that prompts the user to enter a, b,c, d,e and f and displays the result. If ad-bc is 0, report that "The equation has no solution." Enter f 1. 9.0 4.0 3.0 21.0 x is 2.0 and y is 3.0 Enter a, b, c, d, e, f: 1.0 2.0 2.0 4.0 4.0 5.0 Ptin The equation has no solution

Explanation / Answer

******************LInearEquation.cpp*****************

#include<iostream>
using namespace std;
class LinearEquation{
   private:
       float a,b,c,d,e,f;
   public:
       LinearEquation(float p,float q,float r,float s,float t,float u){
           a=p;
           b=q;
           c=r,
           d=s;
           e=t;
           f=u;
       }
       void geta(float p){
           a=p;
       }
       void getb(float p){
           b=p;
       }
       void getc(float p){
           c=p;
       }
       void getd(float p){
           d=p;
       }
       void gete(float p){
           e=p;
       }
       void getf(float p){
           f=p;
       }
       bool solvable(){
           if((a*d)-(b*c)!=0)
               return true;
           else
               false;
       }
       float getX(){
           float x;
           x=((e*d)-(b*f))/((a*d)-(b*c));
           return x;      
           }
       float getY(){
           float y;
           y=((a*f)-(e*c))/((a*d)-(b*c));
           return y;      
           }
      
  
};

int main(){
   float p,q,r,s,t,u;
   cout<<" Enter a,b,c,d,f,e:";
   cin>>p>>q>>r>>s>>t>>u;
   LinearEquation l(p,q,r,s,t,u);
   if(l.solvable()==true){
       cout<<" x is:"<<l.getX();
       cout<<" and y is "<<l.getY();
   }
   else{
       cout<<" The equation has no solution.";
   }

}

**********************OUTPUT************************


Enter a,b,c,d,f,e:9.0
4.0
3.0
-5.0
-6.0
-21.0

x is:-2 and y is 3
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