{"id":24953,"date":"2021-06-21T16:53:51","date_gmt":"2021-06-21T11:23:51","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=24953"},"modified":"2021-06-21T16:53:51","modified_gmt":"2021-06-21T11:23:51","slug":"ncert-solutions-for-class-9-maths-chapter-4-ex-4-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-4-ex-4-3\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 4 Linear Equations in Two Variables Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise 4.3<\/h2>\n

Question 1.
\nDraw the graph of each of the following linear equations in two variables:
\n(i) x + y = 4
\n(ii) x – y = 2
\n(iii) y = 3x
\n(iv) 3 = 2x + y
\nSolution:
\n(i) We have given the equation
\nx + y = 4
\nTo draw the graph, we need at least two solutions to the equation.
\nPut x = 0 then y = 4 and if x = 4 then y = 0 are solutions of the given equation.
\nSo we use the following table to draw the graph:
\n\"NCERT<\/p>\n

(ii) We have given the equation x – y = 2
\nTo draw the graph, we need at least two solutions to the equation.
\nPut x = 0 then y = -2 and if x = 2 then y = 0 are solutions of the given equation.
\nSo we use the following table to draw the graph:
\n\"NCERT<\/p>\n

(iii) We have given the equation y = 3x
\nTo draw the graph, we need at least two solutions to the equation.
\nPut x = 0 the y = 0, and if x = 1 then y = 3 are the solutions of the given equation.
\nSo we use the following table to draw the graph:
\n\"NCERT<\/p>\n

(iv) We have given the equation 3 = 2x + y
\nTo draw the graph, we need at least two solutions to the equation.
\nPut x = 0 then y = 3 and if x = 1 then y = 1 are the solutions of the given equation.
\nSo we use the table to draw the graph:
\n\"NCERT<\/p>\n

Question 2.
\nGive the equations of two lines passing through (2, 14). How many more such lines are there, and why?
\nSolution:
\nHere (2, 14) is the solution of a linear equation which is satisfied by the co-ordinate of the point (2,14).
\nFor eg. 2x + 10 = y, 3x + 8 = y, 4x + 6 = y and so on.
\nIt is because we know that from a given point there are infinitely many straight lines that pass.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nIf the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a?
\nSolution:
\nWe have given that (3, 4) lies on the graph of the equation.
\nTherefore we put x = 3 and y = 4 in this equation.
\n3 \u00d7 4 = a \u00d7 3 + 7
\nor, 12 = 3a + 7
\nor, 12 – 7 = 3a
\nor, a = \\(\\frac {5}{3}\\)
\nTherefore, the value of a = \\(\\frac {5}{3}\\)<\/p>\n

Question 4.
\nThe taxi fare in a city is as follows:
\nFor the first kilometer, the fare is Rs. 8 and for the subsequent distance, it is Rs. 5 per km. Taking the distance covered as x km and total fare is Rs. y, write a linear equation for this information and draw its graph.
\nSolution:
\nWe have given that the total distance covered is x km and the total fare is Rs. y.
\nBut, we have given that the rate of the first km is Rs. 8.
\nTherefore the fare of the remaining distance (x – 1) km is Rs. 5 per km.
\nTherefore, according to question, we can say
\n(x – 1) \u00d7 5 + 8 = y
\nor, 5x – 5 + 8 = y
\nor, 5x + 3 = y
\nTherefore, 5x + 3 = y is required equation for this information.
\nWe use the following table to represent the graph.
\n\"NCERT<\/p>\n

Question 5.
\nFrom the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.
\n(a) For Fig. 4.6
\n(i) y = x
\n(ii) x + y = 0
\n(iii) y = 2x
\n(iv) 2 + 3y = 7x
\n\"NCERT
\n(b) For Fig. 4.7
\n(i) y = x + 2
\n(ii) y = x – 2
\n(iii) y = -x + 2
\n(iv) x + 2y = 6
\n\"NCERT
\nSolution:
\n(a) In Fig. 4.6 the points on the line are (-1, 1), (0, 0) and (1, -1). By inspection x + y = 0 is the equation corresponding to this graph. We find that the coordinate in each case is equal but an opposite sign of the x coordinate.<\/p>\n

(b) In Fig. 4.7 the points on the line are (0, 2) and (2, 0). By inspection y = -x + 2 is the equation corresponding to this graph. We find that when we put x = 0 then the value of y = 2 and when we put x = 2 then the value of y = 0.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nIf the work done by a body on the application of a constant force is directly proportional to the distance travelled by the body. Express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also real from the graph the work done when the distance travelled by the body is
\n(i) 2 units
\n(ii) 0 units
\nSolution:
\nLet us assume,
\nThe work done by a constant force is y-units and the distance travelled by the body is x-units. Therefore, According to the question.
\nForce y is directly proportional to distance travelled x
\ny \u221d x
\ny = kx (where k is constant)
\nAgain according to the question.
\nWe have given that constant force (k) = 5 units.
\nThe required two-variable equation is y = 5x
\n(i) According to the graph,
\nWhen the distance travelled by the body is 2 units, then the work done is 10 units.
\n(ii) According to the graph,
\nWhen the distance travelled by the body is 0 units, the work done is 0 Units.
\nTo draw the graph of y = 5x we use the following table.
\n\"NCERT<\/p>\n

Question 7.
\nYamini and Fatima, two students of class IX of a school, together contributed \u20b9 100 towards Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data, (you may take their contributions as \u20b9 x and \u20b9 y). Draw the graph of the same.
\nSolution:
\nLet Yamini contributed towards Prime Minister Relief Fund is \u20b9 x and Fatima Contributed \u20b9 y.
\nTherefore, According to question,
\nx + y = 100
\nThe linear equation which satisfies this data is x + y = 100.
\nTo draw the graph we use the following table:
\n\"NCERT<\/p>\n

Question 8.
\nIn countries like the USA and Canada, the temperature is measured in Fahrenheit, whereas in countries like India it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:
\nF = (\\(\\frac {9}{5}\\)) C + 32
\n(i) Draw the graph of the linear equation above using Celsius for the x-axis and Fahrenheit for the y-axis.
\n(ii) If the temperature is 30\u00b0C, what is the temperature in Fahrenheit?
\n(iii) If the temperature is 95\u00b0 F, what is the temperature in Celsius?
\n(iv) If the temperature is 0\u00b0C, what is the temperature in Fahrenheit and if the temperature is 0\u00b0F, what is the temperature in Celsius?
\n(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes find it.
\nSolution:
\n(i) We have given that the linear equation.
\nF = (\\(\\frac {9}{5}\\)) C + 32
\nTo draw the graph, we need at least two solutions to the equation.
\nPut C = 0 then F = 32 and if C = 5 then F = 41 are the solutions of the given equation.
\nSo we use the following table to draw the graph.
\n\"NCERT
\nTaking C on the x-axis and F on the y-axis.<\/p>\n

(ii) We have given that C = 30\u00b0
\nThen by the linear equation F = (\\(\\frac {9}{5}\\)) C + 32
\nPut C = 30
\nF = \\(\\frac {9}{5}\\) \u00d7 30 + 32
\nF = 54 + 32
\n\u2234 F = 86
\nTherefore, If the temperature is 30\u00b0C then the temperature in Fahrenheit is 86\u00b0.<\/p>\n

(iii) We have given that F = 95\u00b0
\nThen by the linear equation
\nF = (\\(\\frac {9}{5}\\)) C + 32
\nPut F = 95\u00b0
\nor, 95 = \\(\\frac {9}{5}\\) C + 32
\nor, 95 – 32 = (\\(\\frac {9}{5}\\)) C
\n63 \u00d7 \\(\\frac {5}{9}\\) = C
\n\u2234 C = 35\u00b0
\nThe temperature of 95\u00b0F is 35\u00b0 Celsius.<\/p>\n

\"NCERT<\/p>\n

(iv) We have given that C = 0\u00b0
\nThen by the linear equation
\nF = (\\(\\frac {9}{5}\\)) C + 32
\nPut C = 0\u00b0
\nF= (\\(\\frac {5}{9}\\)) \u00d7 0 + 32
\nor, F = 0 + 32
\nF = 32
\nThe temperature of 0\u00b0C is 32 Fahrenheit.
\nAgain in case II,
\nWe have given that F = 0\u00b0
\nThen by liner equation
\nF = (\\(\\frac {9}{5}\\)) C + 32
\nput F = 0
\n0 = (\\(\\frac {9}{5}\\)) C + 32
\nor, 0 – 32 = (\\(\\frac {9}{5}\\)) C
\nor, -32 \u00d7 \\(\\frac {5}{9}\\) = C
\nor, \\(\\frac {-160}{9}\\) = C
\nC = -17.77
\nor, C = 17.7
\n\u2234 The temperature of 0\u00b0F is -17.7 Celsius.<\/p>\n

\"NCERT<\/p>\n

(v) Yes, there is a temperature which is numerically the same in both Fahrenheit and Celsius.
\nThis is -40\u00b0C
\nAt -40\u00b0C the value of the Fahrenheit scale is also -40\u00b0F.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise 4.3 Question 1. Draw the graph of each of the following linear equations in …<\/p>\n

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NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise 4.3 Question 1. 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