NCERT Class 9 Maths Solutions Chapter - 4 Linear Equations in Two Variables, Ex - 4.2

Ex - 4.2

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

Solution

y = 3x + 5 is a linear equation in two variables and it has infinite solutions. As for every value of x there will be a value of y satisfying 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)    x + y = 9    (iii)    x = 4y     : 

Solution

(i)    2x + y = 7
         For x = 0
    2(0) + y = 7

        y = 7

    So, (0, 7) is a solution of this equation
        For x = 1

    2(1) + y = 7

          y = 5

    So, (1, 5) is a solution of this equation

            For x = -1
     2(-1) + y = 7

              y = 5

      So, (-1, 9) is a solution of this equation

            For x = 2
       2(2) + y = 7
             y = 3

       So (2, 3) is a solution of this equation.

(ii)     + y = 9

             For x = 0

       (o) + y = 9

              y = 9

      So (0, 9) is a solution of this equation
      For x = 1
       (1) + y =9

      y = 9 - 

 So, (1, 9 - ) is a solution of this equation 

      For x = 2

       (2) + y = 9

      y = 9 - 2

       So, (2, 9 -2) is a solution of this equation

       For x = -1

        (-1) + y = 9

                     y = 9 + 

        So, (-1, 9 + ) is a solution of this equation 

(iii)   x = 4y  
        For x = 0
        0 = 4y

         y = 0

        So, (0, 0) is a solution of this equation
        For y = 1
        x = 4(1) = 4
        So, (4, 1) is a solution of this equation
        For y = - 1
        x = 4(-1)
        x = -4
        So, (-4, - 1) is a solution of this equation
        For x = 2
        2 = 4y

        y = 

        So, 

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)         (v) (1,1)

Solution

(i)   (0, 2)
      Putting x = 0, and y = 2 in the L.H.S of given equation
      x - 2y = 0 - (22 )

               = - 4

      As -4 # 4  
      L.H.S # R.H.S
      So (0, 2) is not a solution of this equation.

(ii)  (2, 0)

      Putting x = 2, and y = 0 in the L.H.S of given equation
      x - 2y = 2 - (2  0)

              = 2 

              As 2 # 4
      L.H.S  R.H.S
      So (2, 0) is not a solution of this equation.

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

(iv)   

       Putting x =  and y = 4 in the L.H.S of given equation.

         

                       
      L.H.S  R.H.S
      So   is not a solution of this equation.  

(v)  (1, 1)
      Putting x = 1, and y = 1 in the L.H.S of given equation
      x - 2y  = 1 - 2(1)

                = 1 - 2

                = - 1 

               As -1 # 4
                       L.H.S  R.H.S
      So (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.

Solution

Putting x = 2, and y = 1 in the given equation

2x + 3y = k
2(2) + 3(1) = k
4 + 3 = k
      k = 7 

Ex-4.3

Ex-4.1