NCERT Exemplar Solutions for Class 9 Maths Chapter 4 - Linear Equations in Two Variables PDF

NCERT Exemplar Solutions For Class 9 Maths Chapter 4- Linear Equations In Two Variables Exercise 4.1 Page No: 34 Write the correct answer in each...

NCERT Exemplar Solutions For Class 9 Maths Chapter 4- Linear Equations In Two Variables Exercise 4.1 Page No: 34 Write the correct answer in each of the following: 1. The linear equation 2x – 5y = 7 has A. A unique solution B. Two solutions C. Infinitely many solutions D. No solution Solution: C. Infinitely many solutions Explanation: Expressing y in terms of x in the equation 2x – 5y = 7, we get, 2x – 5y = 7 – 5y = 7 – 2x y = ( 7 – 2x)/– 5 Hence, we can conclude that the value of y will be different for different values of x. Hence, option C is the correct answer. 2. The equation 2x + 5y = 7 has a unique solution, if x, y are: A. Natural numbers B. Positive real numbers C. Real numbers D. Rational numbers Solution: A. Natural numbers Explanation: Consider, 2x + 5y = 7 x 1 y 1 A. (1, 1) is a solution of 2x + 5y = 7 B. If positive real numbers are chosen, 2x + 5y = 7 will have many solutions. C. If real numbers are chosen, 2x + 5y = 7 will have infinite solutions. D. If rational numbers are chosen, 2x + 5y = 7 will have many solutions. Hence, option A is the correct answer. 3. If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is A. 4 B. 6 C. 5 D. 2 Solution: A. 4 Explanation: We know that, (2, 0) = (x, y) Substituting values of x and y in the above equation, we get  NCERT Exemplar Solutions For Class 9 Maths Chapter 4- Linear Equations In Two Variables 2×2 + 3×0 = k k=4 Hence, option A is the correct answer. 4. Any solution of the linear equation 2x + 0y + 9 = 0 in two variables is of the form A. (-9/2 , m) B. (n, -9/2) C. (0, -9/2) D. ( – 9, 0) Solution: A. (-9/2 , m) Explanation: Solving the above equation we get, 2x = -9 x = -9/2 As the coefficient of y is 0, therefore, y can take any value and will not affect our answer. A. x = -9/2 y = any value B. x = n C. x = 0 D. x = -9 Hence, option A is the correct answer. 5. The graph of the linear equation 2x + 3y = 6 cuts the y – axis at the point A. (2, 0) B. (0, 3) C. (3, 0) D. (0, 2) Solution: D. (0, 2) Let 2x + 3y = 6 cut the y-axis at P. therefore at P x-coordinate = 0. Substituting x = 0, we get 2(0) + 3y = 6 3y = 6 y=2 Hence the coordinates are (0, 2). A. (2, 0) is wrong because it has x = 2 B. (0, 3) is wrong because it has y = 3 C. (3, 0) is wrong because it has x = 3 D. (0, 2) is right because it has x = 0 and y = 2 which is equal to the coordinates (0,2) Hence, option D is the correct answer. 6. The equation x = 7, in two variables, can be written as A. 1. x + 1. y = 7 B. 1. x + 0. y = 7  NCERT Exemplar Solutions For Class 9 Maths Chapter 4- Linear Equations In Two Variables C. 0. x + 1. y = 7 D. 0. x + 0. y = 7 Solution: B. 1. x + 0. y = 7 A. Simplifying the equation we get x + y = 7 B. Simplifying the equation we get x + 0y = 7 which is equal to x = 7 C. Simplifying the equation we get y = 7 D. simplifying the equation we get 0x + 0y = 7 which is not possible. Hence, option (B) is the correct answer. 7. Any point on the x – axis is of the form A. (x, y) B. (0, y) C. (x, 0) D. (x, x) Solution: C. (x, 0) Any point on the x-axis has its ordinate 0. So, any point on the x-axis if of the form (x, 0). Hence, option (C) is the correct answer. 8. Any point on the line y = x is of the form A. (a, a) B. (0, a) C. (a, 0) D. (a, – a) Solution: A. (a, a) Any point on the line y = x will have x and y coordinate same. So, any point on the line y = x is of the form (a, a). Hence, option (A) is the correct answer. 9. The equation of x – axis is of the form A. x = 0 B. y = 0 C. x + y = 0 D. x = y Solution: B. y = 0 The equation of x-axis is y = 0, since, x-axis is a parallel to itself at a distance 0 from it. Hence, option (B) is the correct answer.  NCERT Exemplar Solutions For Class 9 Maths Chapter 4- Linear Equations In Two Variables Exercise 4.2 Page No: 37 Write whether the following statements are True or False? Justify your answers: 1. The point (0, 3) lies on the graph of the linear equation 3x + 4y = 12. Solution: True. Justification: We have the equation, 3x + 4y = 12. Substituting the values of x = 0 and y = 3 from the point (0, 3) in the equation, We get, 3(0) + 4(3) = 12 = RHS. Hence, the point (0, 3) lies on the graph of the linear equation 3x + 4y = 12. 2. The graph of the linear equation x + 2y = 7 passes through the point (0, 7). Solution: False. Justification: We have the equation, x + 2y = 7. Substituting the values of x = 0 and y = 7 from the point (0, 7) in the equation, We get, 0 + 2(7) = 14 ≠RHS Hence, the graph of the linear equation x + 2y = 7 passes through the point (0, 7). 3. The graph given below represents the linear equation x + y = 0. Solution: True. Justification:  NCERT Exemplar Solutions For Class 9 Maths Chapter 4- Linear Equations In Two Variables We have the equation, x + y = 0. x+y=0 x=–y from the graph, we get the points ( – 3, 3) and ( – 1, 1), Considering the point ( – 3, 3) x = – 3 and y = 3 Hence, substituting ( – 3, 3) in equation, We get, – 3 = 3 which satisfies the equation x = – y Considering the point ( – 1, 1) x = – 1 and y = 1 Hence, substituting ( – 1, 1) in equation, We get, – 1 = 1 which satisfies the equation x = – y Therefore, the given solution: ( – 3, 3) and ( – 1, 1) satisfies the given equation x = – y. Hence, the given graph represents the linear equation x + y = 0.  NCERT Exemplar Solutions For Class 9 Maths Chapter 4- Linear Equations In Two Variables Exercise 4.3 Page No: 38 1. Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. What do you observe? Solution: According to the question, y = x ------- eq (i) Values of x and y satisfying the equation= x –1 0 1 y –1 0 1 y = -x ------- (ii) Values of x and y satisfying the equation= x –1 0 1 y 1 0 –1 Plotting the graph: From the above graph, We observe that the two lines y = x and y = – x intersect each other at O (0, 0). 2. Determine the point on the graph of the linear equation 2x + 5y = 19 whose ordinate is 1½ times its abscissa. Solution: From the question, we have, 2x + 5y = 19 …(i) According to the question, Ordinate is 1½ times its abscissa ⇒ y = 1½ x = (3/2) x Substituting y = (3/2)x in eq. (i) We get,  NCERT Exemplar Solutions For Class 9 Maths Chapter 4- Linear Equations In Two Variables 2x + 5 (3/2) x = 19 (19/2)x = 19 x=2 Substituting x = 2 in eq. (i) We get 2x + 5y = 19 2(2) + 5y = 19 y = (19 – 4)/5 = 3 Hence, we get x =2 and y = 3 Thus, point (2, 3) is the required solution. 3. Draw the graph of the equation represented by a straight line which is parallel to the x-axis and at 3 units below. Solution: According to the question, We get the linear equation, y=–3 Values of x and y satisfying the equation= x –1 0 1 y –3 –3 –3 Plotting the graph: 4. Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units. Solution: According to the question, We get the linear equation, x + y = 10 We get,  NCERT Exemplar Solutions For Class 9 Maths Chapter 4- Linear Equations In Two Variables x = 10 – y Values of x and y satisfying the equation= x 10 5 0 y 0 5 10 Plotting the graph: 5. Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. Solution: According to the question, A linear equation such that each point on its graph has an ordinate(y) which is 3 times its abscissa(x). So we get, ⇒ y = 3x. Hence, y = 3x is the required linear equation.  NCERT Exemplar Solutions For Class 9 Maths Chapter 4- Linear Equations In Two Variables Exercise 4.4 Page No: 41 1. Show that the points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the graph of the linear equation y = 9x – 7. Solution: We have the equation, y = 9x – 7 For A (1, 2), Substituting the values of (x,y) = (1, 2), We get, 2 = 9(1) – 7 = 9 – 7 = 2 For B (–1, –16), Substituting the values of (x,y) = (–1, –16), We get, –16 = 9(–1) – 7 = – 9 – 7 = – 16 For C (0, –7), Substituting the values of (x,y) = (0, –7), We get, – 7 = 9(0) – 7 = 0 – 7 = – 7 Hence, we find that the points A (1, 2), B (–1, –16) and C (0, –7) satisfies the line y = 9x – 7. Hence, A (1, 2), B (–1, –16) and C (0, –7) are solutions of the linear equation y = 9x – 7 Therefore, points A (1, 2), B (–1, –16), C (0, –7) lie on the graph of linear equation y = 9x – 7. 2. The following observed value of x and y are thought to satisfy a linear equation. Write the linear equation- x 6 –6 y –2 6 Draw the graph using the value of x, y as given in the above table. At what points the graph of the linear equation (i) cuts the X-axis ? (ii) cuts the Y-axis? Solution:  NCERT Exemplar Solutions For Class 9 Maths Chapter 4- Linear Equations In Two Variables We know that, The linear equation of a line is, y = mx + c, where, c is the y-intercept From the graph, We get y-intercept is 2. ⇒ c = 2. Also, from the graph, We get, x1 = 6, y1 = – 2 and x2 = – 6, y2 = 6 We know that, m = slope of the line ∴ we get the linear equation, y = – (2/3)x + 2 Multiplying whole equation by 3, we get, ⇒ 3y = – 2x + 6 ⇒ 2x + 3y – 6 = 0 Thus, the points the graph of the linear equation cuts (i)x-axis Since, the point is on x axis, we have, y = 0. Substituting y = 0 in the equation, 2x + 3y – 6 = 0, We get, 2x + 3×0 – 6 = 0 ⇒ 2x = 6 ⇒x=3 Hence, the point at which the graph cuts x-axis = (3, 0). (ii) y-axis Since, the point is on y axis, we have, x = 0. Substituting x = 0 in the equation, 2x + 3y – 6 = 0, We get, 2×0 + 3y – 6 = 0 ⇒ 3y = 6 ⇒y=2 Hence, the point at which the graph cuts x-axis = (0, 2).  NCERT Exemplar Solutions For Class 9 Maths Chapter 4- Linear Equations In Two Variables 3. Draw the graph of the linear equation 3x + 4y = 6. At what points, the graph cuts X and Y-axis? Solution: According to the question, We get the equation, 3x + 4y = 6. We need at least 2 points on the graph to draw the graph of this equation, Thus, the points the graph cuts (i) x-axis Since, the point is on x axis, we have, y = 0. Substituting y = 0 in the equation, 3x + 4y = 6, We get, 3x + 4×0 = 6 ⇒ 3x = 6 ⇒x=2 Hence, the point at which the graph cuts x-axis = (2, 0). (ii) y-axis Since, the point is on y axis, we have, x = 0. Substituting x = 0 in the equation, 3x + 4y = 6, We get, 3×0 + 4y = 6 ⇒ 4y = 6 ⇒ y = 6/4 ⇒ y = 3/2 ⇒ y = 1.5 Hence, the point at which the graph cuts x-axis = (0, 1.5). Plotting the points (0, 1.5) and (2, 0) on the graph.