Linear Equation in Two Variables

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1 Linear Equation in Two Variables Finish Line & Beond Linear Equations Solution Of A Linear Equation Graph Of A Linear Equation In Two Variables Equations Of Lines Parallel To X-axis And Y- axis

2 (a) Linear Equation in One Variable: - An equation like ax + b 0, i.e. 2x + 1 = 0, is having one variable x. This is known as linear equation in one variable. Such equations have a unique solution. The solution of a linear equation is not affected: - (i) When same number is added (or subtracted) from both sides of the equation. (ii) When we multipl (or divide) both sides of the equation b the same nonzero number. (b) Linear Equations in Two Variables: - An equation like ax + b + c = 0 {where a, b and c are real numbers and at least one of a, b is non-zero}, i.e. 2x = 5. This equation is having two variables x and, that is wh such equations are known as linear equations in two variables.

3 EXERCISE 1 Q1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent the above statement. Sol. Let the cost of a notebook be x and that of a pen be. According to question, The cost of a notebook is twice the cost of a pen x = 2 x - 2 = 0. Q2. Express the following linear equations in the form ax + b + c = 0 and indicate the values of a, b and c in each case: (i) 2x + 3 = 9.35 (ii) x = 0 5 (iii) -2x + 3 = 6 (iv) x = 3 (v) 2x = -5 (vi) 3x + 2 = 0 (vii) 2 = 0 (viii) 5 = 2x Sol. (i) 2x + 3 = x = 0 Comparing the equation with ax + b + c = 0, we have a = 2, b = 3, c = 9.35 (ii) x = 0 Comparing the equation with ax + b + c = 0, we have a = 1, b = - 5 1, c = -10 (iii) -2x + 3 = 6-2x = 0 Comparing the equation with ax + b + c = 0, we have a = -2, b = 3, c = -6 (iv) x = 3 x - 3 = 0 Comparing the equation with ax + b + c = 0, we have a = 1, b = -3, c = 0 (v) 2x = -5 2 x + 5 = 0

4 Comparing the equation with ax + b + c = 0, we have a = 2, b = 5, c = 0 (vi) 3x + 2 = 0 3 x = 0 Comparing the equation with ax + b + c = 0, we have a = 3, b = 0, c = 2 (vii) 2 = 0 0 x = 0 Comparing the equation with ax + b + c = 0, we have a = 0, b = 1, c = -2 (viii) 5 = 2x -2x = 0 Comparing the equation with ax + b + c = 0, we have a = -2, b = 0, c = 5

5 EXERCISE 2 Q1. Which one of the following statements is true and wh? = 3x + 5 has (i) (ii) (iii) a unique solution onl two solutions infinitel man solutions. Sol. Since = 3x + 5 is a linear equation in two variables. We know that a linear equation in two variables has infinite solutions. Hence option (iii) is true. Q2. Write four solutions for each of the following equations: (i) 2x + = 7 (ii) π x + = 9 (iii) x = 4. Sol. (i) 2x + = 7 = 7 2x equation (1) When x = 0, Putting the value in equation (1) = 7 2 0, = 7 0, = 7. Putting the value x = 1 in equation (1) = 7 2 1, = 7 2, = 5. Putting the value x = 2 in equation (1) = 7 2 2, = 7 4, = 3. Putting the value x = 3 in equation (1) = 7 2 3, = 7 6, = 1. Hence, four solutions for equation 2x + = 7 are (0,7), (1,5), (2,3), (3,1). (ii) π x + = 9 = 9 π x equation (1) When x = 0, Putting the value in equation (1) = 9 π 0, = 9 0, = 9. Putting the value x = 1 in equation (1) = 9 π 1, = 9 π. Putting the value x = -1 in equation (1) = 9 π -1, = 9 + π. {since -1-1 = +1} Putting the value x = 2 in equation (1)

6 = 9 π 2, = 9 2 π. Hence, four solutions for equation π x + = 9 are (0, 9), (1, 9 π ), (-1, 9 + π ), (2, 9 2 π ). (iii) x = 4 equation (1) When = 0, Putting the value in equation (1) x = 4 0, x = 0. Putting the value = 1 in equation (1) x = 4 1, = 4. Putting the value = -1 in equation (1) x = 4-1, = -4. Putting the value = 2 in equation (1) x = 4 2, = 8. Hence, four solutions for equation x = 4 are (0, 0), (1, 4), (-4,-1), (8, 2). Q3. Check which of the following are solutions of the equation x 2 = 4 and which are not: (i) (0, 2) (ii) (2, 0) (iii) (4, 0) (iv) ( 2, 4 2 ) (v) (1, 1) Sol. (i) The given equation is x 2 = 4 L.H.S = x 2 Putting the value x = 0 and = 2 to verif the solution (0, 2) = = 0 4 = -4 R.H.S. (0, 2) is not a solution of the equation x 2 = 4. (ii) The given equation is x 2 = 4 L.H.S = x 2 Putting the value x = 0 and = 2 to verif the solution (2, 0) = = 2 0 = 2 R.H.S. (2, 0) is not a solution of the equation x 2 = 4. (iii) The given equation is x 2 = 4 L.H.S = x 2 Putting the value x = 0 and = 2 to verif the solution (4, 0) = = 4 0 = 4 = R.H.S. (4, 0) is a solution of the equation x 2 = 4.

7 (iv) The given equation is x 2 = 4 L.H.S = x 2 Putting the value x = 0 and = 2 to verif the solution ( 2, 4 2 ) = = = 7 2 R.H.S. ( 2, 4 2 ) is not a solution of the equation x 2 = 4. (v) The given equation is x 2 = 4 L.H.S = x 2 Putting the value x = 0 and = 2 to verif the solution (1, 1) = = 1 2 = -1 R.H.S. (1, 1) is not a solution of the equation x 2 = 4. Q4. Find the value of k if x = 2, = 1 is a solution of the equation 2x + 3 = k. Sol. The given equation is 2x + 3 = k Putting the given value x = 2 and = 1 in the equation = k = k 7 = k Hence, k = 7 EXERCISE 3

8 Q1. Draw the graph of each of the following linear equation in two variables: (i) x + = 4 (ii) x = 2 (iii) = 3x (iv) 3 = 2x + Sol. (i) The given equation is x + = 4 = 4 x.equation (1) Now, putting the value x = 0 in equation (1) = 4 0 = 4. So the solution is (0, 4) Putting the value x = 1 in equation (1) = 4 1 = 3. So the solution is (1, 3) Putting the value x = 2 in equation (1) = 4 2 = 2. So the solution is (2, 2) So, the table of the different solutions of the equation is x (ii) The given equation is x - = 2 x = 2 +.equation (1)

9 Now, putting the value = 0 in equation (1) x = = 2. So the solution is (2, 0) Putting the value = 1 in equation (1) x = = 3. So the solution is (3, 1) Putting the value = 2 in equation (1) x = = 4. So the solution is (4, 2) So, the table of the different solutions of the equation is x x-= x (iii) The given equation is = 3x = 3x.equation (1) Now, putting the value x = 0 in equation (1) = 3 0 = 0. So the solution is (0, 0) Putting the value x = 1 in equation (1) = 3 1 = 3. So the solution is (1, 3)

10 Putting the value x = 2 in equation (1) = 3 2 = 6. So the solution is (2, 6) So, the table of the different solutions of the equation is x =3x x (iii) The given equation is 3 = 2 x + 2 x + = 3 = 3 2x.equation (1) Now, putting the value x = 0 in equation (1) = = 3 0 = 3. So the solution is (0, 3) Putting the value x = 1 in equation (1) = = 3 2 = 1. So the solution is (1, 1)

11 Putting the value x = 2 in equation (1) = = 3 4 = -1. So the solution is (2, -1) So, the table of the different solutions of the equation is x =2x x Q2. Give the equations of two lines passing through (2, 14). How man more such lines are there, and wh? Sol. Since the given solution is (2, 14) Therefore, x = 2 and = 14 One equation is x + = = 16 x + = 16 Second equation is x = 2 14 = -12 x - = -12 Third equation is = 7x

12 0 = 7x 7x - = 0 In fact we can find infinite equations because through one point infinite lines pass. Q3. If the point (3, 4) lies on the graph of the equation 3 = ax + 7, find the value of a? Sol. The given equation is 3 = ax + 7.equation (1) According to problem, point (3, 4) lie on it. So, putting the value x = 3 and = 4 in equation (1) 3 4 = a = 3a = 3a 5 = 3a 5 = a 3 a = 3 5 Now putting the value of a in equation (1) 3 = 3 5 x + 7 Now for x = 3 and = 4 L.H.S = 3 = 3 4 = 12 R.H.S = 3 5 x + 7 = = = 12 Hence, L.H.S = R.H.S Or, 3 = 3 5 x + 7. Q4. The taxi fare in a cit is as follows: For the first kilometer, the fare is Rs. 8 and for the subsequent distance it is Rs. 5 per kilometer. Taking the distance covered as x km and total fares as Rs., write a linear equation for this information, and draw its graph.

13 Sol. Given, Taxi fare for first kilometer = Rs. 8 Taxi fare for subsequent distance = Rs. 5 Total distance covered = x Total fare = Since the fare for first kilometer = Rs. 8 According to problem, Fare for (x 1) kilometer = 5(x-1) So, the total fare = 5(x-1) + 8 = 5(x-1) + 8 = 5x = 5x + 3 Hence, = 5x + 3 is the required linear equation. Now the equation is = 5x + 3.equation (1) Now, putting the value x = 0 in equation (1) = = = 3 So the solution is (0, 3) Putting the value x = 1 in equation (1) = = = 8. So the solution is (1, 8) Putting the value x = 2 in equation (1) = = = 13. So the solution is (2, 13) So, the table of the different solutions of the equation is

14 x =5x x Q5. From the choices given below, choose the equation whose graphs are given in Fig. 1 and Fig. 2. (i) = x (i) = x + 2 (ii) x + = 0 (ii) = x 2 (iii) = 2x (iii) = -x + 2 (iv) = 7x (iv) x + 2 = 6

15 x x Sol. From the given Fig. 1, the solutions of the equation are (-1, 1), (0, 0) and (1, -1) Therefore the equation which satisfies these solutions is the correct equation. Equation (ii) x + = 0, satisfies these solutions. Proof: Putting the value x = -1 and = 1 in the equation x + = 0

16 L.H.S = x + = = 0 = R.H.S Putting the value x = 0 and = 0 L.H.S = x + = = 0 = R.H.S Putting the value x = 1 and = -1 L.H.S = x + = 1 + (-1) = 1 1 = 0 = R.H.S Hence, option (ii) x + = 0 is correct. From the given Fig. 2, the solutions of the equation are (-1, 3), (0, 2) and (2, 0) Therefore the equation which satisfies these solutions is the correct equation. Equation (i) = -x + 2, satisfies these solutions. Proof: Putting the value x = -1 and = 3 in the equation = -x + 2 x + = 2 L.H.S = x + = = 2 = R.H.S Putting the value x = 0 and = 2 L.H.S = x + = = 2 = R.H.S Putting the value x = 2 and = 0 L.H.S = x + = = 2 = R.H.S Hence, option (i) = -x + 2 is correct. Q6. The work done b a bod on application of a constant force is directl proportional to the distance travelled b the bod. Express this in the form of an equation in two variables and draw the graph of the same b taking the constant force is 5 units. Read from the graph the work done when the distance traveled b the bod is (i) 2 units (ii) 0 units Sol. Let x units be the distance traveled b the bod and units be the work done b constant force. According to problem, = 5x.equation (1) Now, putting the value x = 0 in equation (1)

17 = 5 0 = 0. So the solution is (0, 0) Putting the value x = 1 in equation (1) = 5 1 = 5. So the solution is (1, 5) Putting the value x = 2 in equation (1) = 5 2 = 10. So the solution is (2, 10) So, the table of the different solutions of the equation is x =5x x (i) When x = 2 units (distance) Putting the value x in equation (1) = 5x = 5 2 = 10. Hence, Work done = 10. (ii) When x = 0 units (distance) Putting the value x in equation (1)

18 = 5x = 5 0 = 0. Hence, Work done = 0. Q7. Yamini and Fatima, two students of class IX of a school, together contributed Rs. 100 towards the Prime Minister s Relief Fund, to help the earthquake victims. Write a linear equation which this data satisfies. (You ma take their contributions as Rs. x and Rs. ). Draw the graph of the same. Sol. Let the contribution of Yamini be x and that of Fatima be. According to problem, x + = 100..(1) Now, putting the value x = 0 in equation (1) 0 + = 100 = 100. So the solution is (0, 100) Putting the value x = 50 in equation (1) 50 + = 100, = = 50. So the solution is (50, 50) Putting the value x = 100 in equation (1) = 100, = = 0. So the solution is (100, 0) So, the table of the different solutions of the equation is x

19 120 x+= x Q8. In countries like the USA and Canada, 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. F = ( 5 9 ) C + 32 (i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for -axis. (ii) If the temperature is 30 C, what is the temperature in Fahrenheit? (iii) If the temperature is 95 F, what is the temperature in Celsius? (iv) If the temperature is 0 C, what is the temperature in Fahrenheit and if the temperature is 0 F, what is the temperature in Celsius? (v) Is there a temperature which is numericall the same in both Fahrenheit and Celsius? If es, find it. Sol. Given equation is F = ( 5 9 ) C + 32 (i) We have to take Celsius along x-axis and Fahrenheit along -axis Let C be x and F be So the equation will be = ( 5 9 ) x + 32 equation (1) Now, putting the value x = 0 in equation (1)

20 = ( 5 9 ) x = = 32. So the solution is (0, 32) Putting the value x = 5 in equation (1) = ( 5 9 ) x = = 41. So the solution is (5, 41) Putting the value x = -5 in equation (1) = ( 5 9 ) x (-5) + 32 = = 23. So the solution is (-5, 23) So, the table of the different solutions of the equation is x =(9/5)x x (ii) When C = 30, F = = = = 86 (iii) When F = 95,

21 95 = 5 9 C = 5 9 C 63 = 5 9 C 5 9 C = 63 C = = 7 5 = 35. (iv) When C = 0, F = = = 32. (v) When x F = x C F = ( 5 9 ) C + 32 x = ( 5 9 ) x + 32 x x = 32 5x 9x 5 = 32 4x 5 = 32-4x = 32 5 x = = -8 5 = - 40 C.

22 EXERCISE 4 Q1. Give the geometric representations of = 3 as an equation. (i) in one variable (ii) in two variables. Sol. (i) = (ii) = 3 0x + = 3 0x + 3 = 0 which is in fact = 3 It is a line parallel to x-axis at a positive distance of 3 from it. We have two solution for it. i.e. (0, 3), (1, 3). Y (0, 3) A B (1, 3) X X -2 Y Q2. Give the geometric representation of 2x + 9 = 0 as and equation,

23 (i) in one variable (ii) in two variables. Sol. (i) 2x + 9 = 0 2x = -9 x = = (iii) Given equation is X x + 9 = X 2x = 0 {we know that it is actuall 2x + 9 = 0} x = = It is line parallel to -axis at a negative distance we have the two points ling it, the points are A(-4.5, 0), B(-4.5, 2). Y B (-4.5, 2) 2 A (-4.5, 0) 1 X X -2 Y

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Downloaded from Exercise 4.1 Question 1: The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be Rs x and that of a pen