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NCERT Solutions 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4
If you are interested in reading NCERT Solutions for Chapter 3 Class 10 Linear Equations then you are at the right frame of mind. You will get all subjects correct and to the point CBSE NCERT Solutions. manabadi.co.in provides these strategies that will aid you a great deal in scoring good marks in the exams.
Exercise 3.4 1. Solve the following pair of linear equations by the elimination method and the substitution method: (i) x + y =5 and 2x –3y = 4 (ii) 3x + 4y = 10 and 2x – 2y = 2 (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 (iv) x/2 + 2y/3 = - 1 and x – y/3 = 3 Answer (i) x + y =5 and 2x –3y = 4 By elimination method x + y =5 ... (i) 2x –3y = 4 ... (ii) Multiplying equation (i) by (ii), we get 2x + 2y = 10 ... (iii) 2x –3y = 4 ... (ii) Subtracting equation (ii) from equation (iii), we get 5y = 6 y = 6/5 Putting the value in equation (i), we get x = 5 - (6/5) = 19/5 Hence, x = 19/5 and y = 6/5 By substitution methodx + y = 5 ... (i) Subtracting y both side, we get x = 5 - y ... (iv) Putting the value of x in equation (ii) we get 2(5 – y) – 3y = 4 -5y = - 6 y = -6/-5 = 6/5 Putting the value of y in equation (iv) we get x = 5 – 6/5 x = 19/5 Hence, x = 19/5 and y = 6/5 again (ii) 3x + 4y = 10 and 2x – 2y = 2 By elimination method 3x + 4y = 10 .... (i) 2x – 2y = 2 ... (ii) Multiplying equation (ii) by 2, we get 4x – 4y = 4 ... (iii) 3x + 4y = 10 ... (i) Adding equation (i) and (iii), we get 7x + 0 = 14 Dividing both side by 7, we get x = 14/7 = 2 Putting in equation (i), we get 3x + 4y = 10 3(2) + 4y = 10 6 + 4y = 10 4y = 10 – 6 4y = 4 y = 4/4 = 1 Hence, answer is x = 2, y = 1 By substitution method 3x + 4y = 10 ... (i) Subtract 3x both side, we get 4y = 10 – 3x Divide by 4 we get y = (10 - 3x )/4 Putting this value in equation (ii), we get 2x – 2y = 2 ... (i) 2x – 2(10 - 3x )/4) = 2 Multiply by 4 we get 8x - 2(10 – 3x) = 8 8x - 20 + 6x = 8 14x = 28 x = 28/14 = 2 y = (10 - 3x)/4 y = 4/4 = 1 Hence, answer is x = 2, y = 1 again. (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 By elimination method 3x – 5y – 4 = 0 3x – 5y = 4 ...(i) 9x = 2y + 7 9x – 2y = 7 ... (ii) Multiplying equation (i) by 3, we get 9 x – 15 y = 11 ... (iii) 9x – 2y = 7 ... (ii) Subtracting equation (ii) from equation (iii), we get -13y = 5 y = -5/13 Putting value in equation (i), we get 3x – 5y = 4 ... (i) 3x - 5(-5/13) = 4 Multiplying by 13 we get 39x + 25 = 52 39x = 27 x =27/39 = 9/13 Hence our answer is x = 9/13 and y = - 5/13 By substitution method 3x – 5y = 4 ... (i) Adding 5y both side we get 3x = 4 + 5y Dividing by 3 we get x = (4 + 5y )/3 ... (iv) Putting this value in equation (ii) we get 9x – 2y = 7 ... (ii) 9 ((4 + 5y )/3) – 2y = 7 Solve it we get 3(4 + 5y ) – 2y = 7 12 + 15y – 2y = 7 13y = - 5 y = -5/13 Hence we get x = 9/13 and y = - 5/13 again. (iv) x/2 + 2y/3 = - 1 and x – y/3 = 3 By elimination method x/2 + 2y/3 = -1 ... (i) x – y/3 = 3 ... (ii) Multiplying equation (i) by 2, we get x + 4y/3 = - 2 ... (iii) x – y/3 = 3 ... (ii) Subtracting equation (ii) from equation (iii), we get 5y/3 = -5 Dividing by 5 and multiplying by 3, we get y = -15/5 y = - 3 Putting this value in equation (ii), we get x – y/3 = 3 ... (ii) x – (-3)/3 = 3 x + 1 = 3 x = 2 Hence our answer is x = 2 and y = -3. By substitution method x – y/3 = 3 ... (ii) Add y/3 both side, we get x = 3 + y/3 ... (iv) Putting this value in equation (i) we get x/2 + 2y/3 = - 1 ... (i) (3+ y/3)/2 + 2y/3 = -1 3/2 + y/6 + 2y/3 = - 1 Multiplying by 6, we get 9 + y + 4y = - 6 5y = -15 y = - 3 Hence our answer is x = 2 and y = -3.Post your comments
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