Linear Equations in Two Variables: 9 Math
NCERT Exercise 4.2: part:2: 9th math
We are learning to find the solutions for linear equations in two variables in this NCERT exercise 4.2 class nine math. In question (3) the solution for the given linear equation in two variables is to find from the given five options, and it is also to point out which solution is correct and which is not. In question (4) of the exercise using the given solution of the given linear equation in two variables, the value of the constant term is to find.
Solution of NCERT Exercise 4.2 questions (3) and (4)
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, 42)
(v) (1, 1)
Solution
Given, linear equation in two variables is x – 2y = 4
By taking the given option (i) (0, 2)
In the option (i) x = 0 and y = 2
Here, LHS = x – 2y
Thus, after substituting the values x = 0 and y = 2 in the LHS of the given linear equation in two variable, x – 2y = 4, we get
0 – 2 × 2
= 0 – 4
= – 4 which is not equaal to the RHS 4
This means – 4 ≠ 4
Thus, option (i) (0, 2) is not the correct solution for the given linear equation in two variables.
By taking the given option (ii) (2, 0)
In the option (ii) x = 2 and y = 0
Here, LHS = x – 2y
Thus, after substituting the values x = 2 and y = 0 in the LHS of the given linear equation in two variable, x – 2y = 4, we get
2 – 2 × 0
= 2 – 0 = 2
⇒ LHS = 2
Clearly, LHS (2) ≠ RHS (4)
Thus, option (ii) (2, 0) is not the correct solution for the given linear equation in two variables.
By taking the given option (iii) (4, 0)
Here, in the option (iii) x = 4 and y = 0
And, in the question, LHS = x – 2y
Thus, after substituting the values x = 4 and y = 0 in the LHS of the given linear equation in two variable, x – 2y = 4, we get
4 – 2 × 0
= 4 – 0 = 4
⇒ LHS = 4
Clearly, LHS (4) = RHS (4)
Thus, option (iii) (4, 0) is the correct solution for the given linear equation in two variables.
By taking the given option (iv) (2, 42)
Here, in the option (iv) x = 2 and y = 42
And, in the given question, LHS = x – 2y
Thus, after substituting the values x = 2 and y = 42 in the LHS of the given linear equation in two variable, x – 2y = 4, we get
2 – 2 × 42
= 2 – 0 = 2
⇒ LHS = 2
Clearly, LHS (2) ≠ RHS (4)
Thus, option (iv) (2, 42) is not the correct solution for the given linear equation in two variables.
By taking the given option (v) (1, 1)
Here, in the option (v) x = 1 and y = 1
And, in the question, LHS = x – 2y
Thus, after substituting the values x = 1 and y = 1 in the LHS of the given linear equation in two variable, x – 2y = 4, we get
1 – 1 × 0
= 1 – 0 = 1
⇒ LHS = 1
Clearly, LHS (1) ≠ RHS (4)
Thus, option (v) (1, 1) is not the correct solution for the given linear equation in two variables.
Thus, for the given linear equation in two variables the only correct solution is option (iii) (4, 0) and others given are not Answer
Question (4) Find the value of k, if x = 2, y =1 is a solution of the equation 2x + 3y = k
Solution
Given, linear equation in two variables is 2x + 3y = k
And also given, x = 2, y = 1
Thus, the value of k = ?
After substituting the given values of x = 2 and y = 1 in the given linear equation in two variables 2x + 3y = k, we get
2 × 2 + 3 × 1 = k
⇒ 4 + 3 = k
⇒ 7 = k
⇒ k = 7
Therefore, for the given values of x, and y for the given linear equation in two variables, the value of k = 7
Thus, k = 7 Answer
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