Exercise 4.2 class 9 Maths | Linear equations in two variables Class 9 maths

Exercise 4.2 class 9 Maths | Linear equations in two variables Class 9 maths Exercise 4.3 class 9 by ‪@rajansir07‬ #Class9Maths #LinearEquations #Exercise4_2 #MathsWithRajanSir #CBSEMaths #MathsTutorial #NCERTSolutions #Class9NCERT #MathsExamPrep #LearnMaths Then, πx +y = 9 (π×1)+y = 9 π+y = 9 y = 9-π (1, 9-π) Let y = 0 Then, πx+y = 9 πx+0 = 9 πx = 9 x = 9/π (9/π,0) Let x = -1 Then, πx + y = 9 (π×-1) + y = 9 -π+y = 9 y = 9+π (-1,9+π) The solutions are (0,9), (1,9-π), (9/π,0), (-1,9+π). x -2y = 4 ⟹ 2-(2×0) = 4 ⟹ 2 -0 = 4 But, 2 ≠ 4 (2, 0) is not a solution of the equation x-2y = 4. (iii) (4, 0) Solution: (x,y) = (4, 0) Here, x= 4 and y=0 Substituting the values of x and y in the equation x -2y = 4, we get, x–2y = 4 ⟹ 4 – 2×0 = 4 ⟹ 4-0 = 4 ⟹ 4 = 4 (4, 0) is a solution of the equation x–2y = 4. (iv) (√2, 4√2) Solution: (x,y) = (√2,4√2) Here, x = √2 and y = 4√2 Substituting the values of x and y in the equation x–2y = 4, we get, x –2y = 4 ⟹ √2-(2×4√2) = 4 √2-8√2 = 4 But, -7√2 ≠ 4 (√2,4√2) is not a solution of the equation x–2y = 4. (v) (1, 1) Solution: (x,y) = (1, 1) Here, x= 1 and y= 1 Substituting the values of x and y in the equation x–2y = 4, we get, x –2y = 4 ⟹ 1 -(2×1) = 4 ⟹ 1-2 = 4 But, -1 ≠ 4 (1, 1) is not a solution of the equation x–2y = 4 4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x+3y = k. Solution: The given equation is 2x+3y = k According to the question, x = 2 and y = 1. Now, substituting the values of x and y in the equation 2x+3y = k, We get, (2×2)+(3×1) = k ⟹ 4+3 = k ⟹ 7 = k k = 7 The value of k, if x = 2, y = 1 is a solution of the equation 2x+3y = k, is 7.