Question-1 :-  Which one of the following options is true, and why? y = 3x + 5 has
(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions

  y = 3x + 5 is a linear equation in two variables and it has infinite possible solutions. 
  As for every value of x, there will be a value of y satisfying the above equation and vice-versa.   
  Hence, the correct answer is (iii).
    

Question-2 :-  Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y

(i) 2x + y = 7   
  For x = 0,   
  2(0) + y = 7   
  y = 7   
  Therefore, (0, 7) is a solution of this equation.   
    
  For x = 1,  
  2(1) + y = 7  
  y = 5   
  Therefore, (1, 5) is a solution of this equation.  
     
  For x = −1,   
  2(−1) + y = 7   
  y = 9   
  Therefore, (−1, 9) is a solution of this equation.  
     
  For x = 2,   
  2(2) + y = 7   
  y = 3   
  Therefore, (2, 3) is a solution of this equation.  
    

 
(ii) πx + y = 9  
  For x = 0, π(0)  
  0 + y = 9  
  y = 9   
  Therefore, (0, 9) is a solution of this equation.  

  For x = 1, 
  π(1) + y = 9  
  y = 9 − π   
  Therefore, (1, 9 − π) is a solution of this equation.
      
  For x = 2, 
  π(2) + y = 9   
  y = 9 − 2π   
  Therefore, (2, 9 −2π) is a solution of this equation.  

  For x = −1, 
  π(−1) + y = 9   
  y = 9 + π   
  (−1, 9 + π) is a solution of this equation.  
  

   
(iii) x = 4y   
  For x = 0,   
  0 = 4y   
  y = 0   
  Therefore, (0, 0) is a solution of this equation.
      
  For y = 1, 
  x = 4(1) = 4   
  Therefore, (4, 1) is a solution of this equation.  

  For y = −1, 
  x = 4(−1)   
  x = −4   
  Therefore, (−4, −1) is a solution of this equation.   

  For x = 2,
  2 = 4y
  y = 2/4
  y = 1/2
  Therefore, (2, 1/2) is a solution of this equation.
    

Question-3 :-  Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0, 2)   (ii) (2, 0)  (iii) (4, 0)  (iv) (√2, 4√2)   (v) (1, 1)

(i) (0, 2)   
  Putting x = 0 and y = 2 in the L.H.S of the given equation, 
  x − 2y = 0 − 2 x 2 = − 4 ≠ 4   
  L.H.S ≠ R.H.S   
  Therefore, (0, 2) is not a solution of this equation.   

(ii) (2, 0)   
  Putting x = 2 and y = 0 in the L.H.S of the given equation, 
  x − 2y = 2 − 2 × 0 = 2 ≠ 4    
  L.H.S ≠ R.H.S   
  Therefore, (2, 0) is not a solution of this equation.   

(iii) (4, 0)   
  Putting x = 4 and y = 0 in the L.H.S of the given equation, x  
  − 2y = 4 − 2(0)   
  = 4 = R.H.S   
  L.H.S. = R.H.S.
  Therefore, (4, 0) is a solution of this equation.
   

     
(iv) (√2, 4√2)
  Putting x = √2 and y = 4√2 in the L.H.S. of the given equation,
  x - 2y = √2 - 2(4√2) 
  = √2 - 8√2 = -7√2 ≠ 4
  L.H.S. ≠ R.H.S.
  Therefore, (√2, 4√2) is not a solution of given equation.
    

    
(v) (1, 1) is not a solution of this equation.   
  Putting x = 1 and y = 1 in the L.H.S of the given equation, 
  x − 2y = 1 − 2(1) = 1 − 2 = − 1 ≠ 4   
  L.H.S ≠ R.H.S   
  Therefore, (1, 1) is not a solution of this equation.  
    

Question-4 :-  Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

  Putting x = 2 and y = 1 in the given equation,   
  2x + 3y = k   
  2(2) + 3(1) = k  
  4 + 3 = k    
  k = 7   
  Therefore, the value of k is 7.