lass IX Math NCERT Solution for Linear Equations in Two Variables 4.2

EXERCISE: 4.2
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
Sol: Option (iii) is true because a linear equation has an infinitely many solutions.
2.   Write four solutions for each of the following equations:
      (i) 2x + y = 7      (ii) πx + y = 9      (iii) x = 4y
Sol: (i) 2x + y = 7
      When x = 0, 2(0) + y = 7
      ⇒                      0 + y = 7
      ⇒                      y =7
      ∴ Solution is (0, 7).
      When x = 1, 2(1) + y = 7
      ⇒                      y = 7 – 2
      ⇒                      y = 5
      ∴ Solution is (1, 5).
      When x = 2, 2(2) + y = 7
      ⇒                      y = 7 – 4
      ⇒                      y = 3
      ∴ Solution is (2, 3).
      When x = 3, 2(3) + y = 7
      ⇒                      y = 7 – 6
      ⇒                      y = 1
      ∴ Solution is (3, 1).
(ii) πx + y = 9
      When x = 0 π(0) + y = 9
      ⇒                      y = 9 – 0
      ⇒                      y = 9
      ∴ Solution is (0, 9).
      When × = 1, π(1) + y = 9
      ⇒                      y = 9 – π
      ∴ Solution is {1, (9 – π)}
      When x = 2, π(2) + y = 9
      ⇒                      y = 9 – 2π
      ∴ Solution is {2, (9 – 2π)}
      When × = –1, π(–1) + y = 9
      ⇒                      – π + y = 9
      ⇒                      y = 9 + π
      ∴ Solution is {–1, (9 + π)}
(iii) x = 4y
      When x = 0, 4y = 0
      ⇒                      y = 0
      ∴ Solution is (0, 0).
      When x = 1, 4y = 1
      ⇒                      y = 0
      ∴ Solution is (0, 0)
      When x = 4, 4y = 4
      ⇒                      
      
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)
Sol: (i) (0, 2) means x = 0 and y = 2
      Putting x = 0 and y = 2 in x – 2y = 4, we have
                  L.H.S. = 0 – 2(2) = –4
      But       R.H.S. = 4
                  L.H.S. ≠ R.H.S.
      ∴ x = 0, y = 0 is not a solution.
(ii) (2, 0) means x = 2 and y = 0
      ∴ Putting x = 2 and y = 0 in x – 2y = 4, we get
                  L.H.S. = 2 – 2(0) = 2 – 0 = 2
      But       R.H.S. = 4
                  L.H.S. ≠ R.H.S.
      ∴ (2, 0) is not a solution.
(iii) (4, 0) means x = 4 and y = 0
      Putting x = 4 and y = 0 in x – 2y = 4, we get
                  L.H.S. = 4 – 2(0) = 4 – 0 = 4
      But       R.H.S. = 4
                  L.H.S. ≠ R.H.S.
      ∴ (4, 0) is a solution.
(v) (1, 1) means x = 1 and y = 1
      Putting x = 1 and y = 1 in x – 2y = 4, we get
      L.H.S. = 1 – 2(1) = 1 – 2 = –1
      But       R.H.S. = 4
      ⇒                      L.H.S ≠ R.H.S.
      ∴ (1, 1) is not a solution.


4.   Find the value of k, if x = 2, y = 1 is a solution fo the equation 2x + 3y = k.
Sol: We have 2x + 3y = k
      Putting x = 2 and y = 1 in 2x + 3y = k, we get
          2(2) + 3(1) = k
      ⇒                      4 + 3 = k
      ⇒                      7 = k
          Thus, the required value of k = 7.

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