$${7x – 15y =  2 ......... (1) }$$                                                          
$${x + 2y =  3 .............(2)}$$   

Solution :
Step 1 : Find the value of one variable, say x in terms of the other variable,
         i.e., y from either equation, whichever is convenient.


We pick either of the equations and write one variable in terms of the other.
	Let us consider the Equation (2) :
	${x + 2y =  3 .............(2)}$   
	and write it as 
	${x=3-2y  .................(3)}$
	
Step 2 :Substitute this value of x in the other equation, and reduce it to an equation in one variable,
          i.e., in terms of y, which can be solved. 

Substitute the value of x in Equation (1). We get
      ${7(3 – 2y) – 15y =  2}$ i.e.,    ${21 – 14y – 15y =  2}$ 
	  i.e., ${– 29y = - 19}$
      Therefore,  ${y=19/29}$
	  
Step 3 : Substitute the value of y (or x) obtained in Step 2 in the equation no 3 obtained in
          Step 1 to obtain the value of the other variable.

Substituting this value of y in Equation (3), we get

        ${x=3-2(19/29) = 49/29}$
		
Therefore the solution is ${x=49/29}$ and ${y=19/29}$

Solution : Let s and t be the current ages (in years) of Aftab and his daughter, respectively.
			 
		first pair of linear equations is represented as follows.
		Seven years ago, I was seven times as old as you were then:
		7 years ago , there ages were ${(s-7)}$ and ${(t-7)}$ respectively.So,
		${s-7=7(t-7)}$
		${s-7=7t-49}$
		${s-7t+42=0 .........................(1)}$

		Second pair of equation is respresented as follows:
		Three years from now, I shall be three times as old as you will be:
		3 years from now , there ages will be ${s+3}$ and ${t+3}$ respectively.So,
		${s + 3 = 3 (t + 3)}$,
		i.e., ${s – 3t -6=0 .......................(2) }$ 

	Now apply the three steps of substitution methods.
Step 1 : Find the value of one variable terms of the other variable
		Lets us consider equation no (1)
		${s-7t+42=0}$
		${s=7t-42  ..........................(3)}$

Step 2 :Substitute this value of s in the other equation, and reduce it to an equation in one variable,
          i.e., in terms of t. 
		 ${s – 3t -6=0}$
         ${(7t-42) – 3t -6=0}$	
		 ${4t-48=0}$
		 ${t=12}$
		 
Step 3: Substituting  value of t obtained in Step 2 in Equation (3) obtained in Step 1 , 
        to get value of other variable.
		${s=7t-49}$
		${s=7(12)-49}$
		${s=42 }$
		
Answer:So, Aftab and his daughter are 42 and 12 years old, respectively. 

Solution : 
    Let the cost of each pencil be Rs ${x}$ and cost of each eraser be Rs ${y}$.
	First linear equation:
	the cost of 2 pencils and 3 erasers is Rs 9.
	${2x+3y=9 .................(1)}$
	
	Second linear equation:
	cost of 4 pencils and 6 erasers is Rs 18.
	${4x+6y=18 .................(2)}$
	
	
	Now apply the three steps of substitution methods.
Step 1 : Find the value of one variable terms of the other variable
        Take equation (1)
		${2x=9-3y ..................(3)}$
		Multiply by 2, hence
		${4x=18-6y}$

Step 2 :Substitute this value of one variable in the other equation, and reduce it to an equation in one variable,
          i.e., in terms of y. 
         ${4x+6y=18}$
		 ${18-6y+6y=18}$
		 ${18=18}$
		 
		 This statement is true for all values of y. However, we do not get a specific value of y as a solution.
		 Therefore, we cannot obtain a specific value of x. 
		 
Step 3: Substituting  value of one variable obtained in Step 2 in Equation (3) obtained in Step 1 , 
        to get value of other variable.
         
		 We cannot obtain a specific value of x.This situation has arisen because both the given equations are the same.
		 Therefore, Equations (1) and (2) have  infinitely  many  solutions
		
Answer:Equations (1) and (2) have  infinitely  many  solutions  

Soultion: The pair of linear equations formed were:

			${x + 2y – 4 = 0}$                           (1)
			
			${2x + 4y – 12 = 0}$                          (2)
			
			We express ${x}$ in terms of ${y}$ from Equation (1) to get
			
			${x =  4 – 2y}$                             (3)
			
			Substituting x in equation (2)
			
			${2(4-2y) + 4y - 12 =0}$
			
			${8-4y+4y-12=0}$
			
			${-4=0}$, which is a false statement.
			Therefore, the equations do not have a common solution. So, the two rails will not cross each other.