${a_1x +b_1 y + c_1  =  0}$                                                             (1)
         ${a_2x + b_2 y + c_2  =  0}$                                                             (2)

 

To obtain the values of x and y as shown above, follow the following steps:
 
Step 1 : Multiply Equation (1) by ${b_2}$ and Equation (2) by ${b_1}$ ,  to get 
		${b_2a_1x + b_2b_1 y + b_2c_1  =  0}$                                                             (3)
		${b_1a_2x + b_1b_2 y + b_1c_2  =  0}$                                                             (4)
		
		
Step 2 : Subtracting Equation (4) from (3), we get:
       ${(b_2a_1  – b_1a_2) x + (b_2b_1  – b_1b_2 ) y + (b_2c_1– b_1c_2) = 0}$

	   i.e., ${(a_1b_2   – a_2b_1) x =  b_1c_2  –   b_2c_1}$ 
	   
	   ${ x=(b_1c_2  –   b_2c_1)/(a_1b_2   – a_2b_1) }$    provided  ${a_1b_2   – a_2b_1 ≠ 0}$           (5) 
	   

Step 3 : Substituting this value of ${x}$ in (1) or (2), we get	
             ${y =   (c_1a_2  − c_2 a_1)/(a_1b_2  − a_2b_1)}$                                           (6)  

Now, two cases arise :
Case 1 :  ${a_1b_2– a_2b_1 ≠ 0}$. In this case ${a_1/a_2 ≠ b_1/b_2}$ , 
        Then the pair of linear equations has a unique  solution.
		
Case 2 :  ${a_1b_2– a_2b_1 = 0}$. In this case ${a_1/a_2 = b_1/b_2 = k}$,

          then ${a_1=ka_2}$ ; ${b_1=kb_2}$

        Substituting the values of ${a}$ and ${b}$  in the Equation (1), we get

        ${k (a_2 x + b_2 y) + c_1=  0}$.                                                (7)

	It can be observed that the Equations (7) and (2) can both be satisfied only if
		
		${c_1  = k c_2}$, i.e ${c_1/c_2=k}$
		
	If  ${c_1  = k c_2}$ , any solution of Equation (2) will satisfy the Equation (1), and vice verse.
		
	So if, ${a_1/a_2=b_1/b_2=c_1/c_2=k}$ , then are infinitely many solutions to the pair of linear equations given by (1) and (2).

	If ${c_1 ≠ k c_2}$, then any solution of Equation (1) will not satisfy Equation (2) and vice versa. Therefore the pair has no solution.
	
    Note that you can write the solution given by Equations (5) and (6) in the following form : 
	
	$${x/(b_1c_2  − b_2c_1) = y/(c_1a_2   − c_2a_1) = 1/(a_1b_2  − a_2b_1)}$$
	

Solution: Using Cross -Multiplication Method to solve the question.

According to cross-multiplication method,
for any given pairs of linear equation,
		${a_1x +b_1 y + c_1  =  0}$
		${a_2x + b_2 y + c_2  =  0}$ 

${x/(b_1c_2  − b_2c_1) = y/(c_1a_2   − c_2a_1) = 1/(a_1b_2  − a_2b_1)}$

Let us denote the cost of an orange by ₹ x and the cost of an apple by ₹ y. 
Then, the equations formed are :
		${5x + 3y =  35},  i.e.,  {5x + 3y – 35 = 0}$              (1)
		${2x + 4y =  28},  i.e., {2x + 4y – 28 = 0}$                (2)
		
${x/(3(-28)  − 4(-35)) = y/(-35(2)   − (-28)5) = 1/(5(4)  − 2(3))}$

${x=(-84+140)/(20-6)}$ = ${56/14}$ = ${4}$

${y=(-70+140)/(20-6)}$ = ${70/14}$ = ${5}$

Answer:	Thus the cost of orange is ₹ 4 and apple is ₹ 5. 

Solution:
Let the fares from bus stand to Malleswaram be ₹ ${x}$ and fare from bus stand to Yeshwanthpur be ₹ ${y}$. 

Pair of linear equation formed are:
 2 tickets to Malleswaram and 3 tickets to Yeshwanthpur , the total cost is ₹ 46.
		${2x+3y=46; 2x+3y-46=0}$                            (1)
		
3 tickets to Malleswaram and 5 tickets to Yeshwanthpur the total cost is ₹ 74.		
		${3x+5y=74 ; 3x+5y-74=0}$                            (2)

For cross-multiplication method,

	${x/(b_1c_2  − b_2c_1) = y/(c_1a_2   − c_2a_1) = 1/(a_1b_2  − a_2b_1)}$
	
	${x/{(3)(-74) - (5)(-46)} = y/{(-46)(3)-(-74)()2)} = 1/{(2)(5)-(3)(3)}}$
	
	${x/8=y/10=1/1}$
	
	${x=8}$; and ${y=10}$
Answer:Hence, the fare from the bus stand in Bangalore to Malleswaram is ₹ 8 and the fare to
Yeshwanthpur is ₹ 10.

Example: For which values of ${p}$ does the pair of equations given below has unique solution?
${4x + py + 8 =  0}$
${2x + 2y + 2 =  0}$ 

Solution : Here ${a_1  = 4}$, ${a_2= 2}$, ${b_1= p}$, ${b_2= 2}$.
Now for the given pair to have a unique solution : ${a_1/a_2 ≠ b_1/b_2}$
i.e ${4/2≠ p/2}$
i.e.   ${p ≠ 4}$
Therefore, for all values of ${p}$, except 4, the given pair of equations will have a unique solution.

Example: For what values of ${k}$ will the following pair of linear equations have infinitely
 many solutions?
${kx + 3y – (k – 3) =  0}$
${12x + ky – k =  0}$

Solution : 
   Here ${a_1/a_2 =k/12}$ ; ${b_1/b_2 = 3/k}$;   ${c_1/c_2= {(-(k-3))/(-k)}}$
   
   For a pair of linear equations to have infinitely many solutions :
   ${a_1/a_2=b_1/b_2=c_1/c_2}$
   So, ${k/12=3/k=(k-3)/k}$
   ${k/12=3/k}$ which gives ${k^2=36}$ i.e. ${k = ± 6}$
   
   Also, ${3/k=(k-3)/k}$ , 
   which gives ${3k = k^2  – 3k}$, i.e., ${6k = k2}$,  which means ${k = 0}$ or ${k = 6}$