Page 1 Exercise 3C 1. Solve the system of equations by using the method of cross multiplication: x + 2y + 1 = 0, 2x 3y 12 = 0. Sol: The given equations are: 2x 3y Here a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = -3 and c2 = -12 By cross multiplication, we have: = = = = = = x = = 3, y = = -2 Hence, x = 3 and y = -2 is the required solution. 2. Solve the system of equations by using the method of cross multiplication: 3x - 2y + 3 = 0, 4x + 3y 47 = 0 Sol: The given equations are: Page 2 Exercise 3C 1. Solve the system of equations by using the method of cross multiplication: x + 2y + 1 = 0, 2x 3y 12 = 0. Sol: The given equations are: 2x 3y Here a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = -3 and c2 = -12 By cross multiplication, we have: = = = = = = x = = 3, y = = -2 Hence, x = 3 and y = -2 is the required solution. 2. Solve the system of equations by using the method of cross multiplication: 3x - 2y + 3 = 0, 4x + 3y 47 = 0 Sol: The given equations are: 3x - 2y 4x + 3y Here a1 = 3, b1 = -2, c1 = 3, a2 = 4, b2 = 3 and c2 = -47 By cross multiplication, we have: = = = = = = x = = 5, y = = 9 Hence, x = 5 and y = 9 is the required solution. 3. Solve the system of equations by using the method of cross multiplication: 6x - 5y - 16 = 0, 7x - 13y + 10 = 0 Sol: The given equations are: 6x - 5y - 7x - Here a1 = 6, b1 = -5, c1 = -16, a2 = 7, b2 = -13 and c2 = 10 By cross multiplication, we have: = = = = = = x = = 6, y = = 4 Hence, x = 6 and y = 4 is the required solution. 4. Solve the system of equations by using the method of cross multiplication: 3x + 2y + 25 = 0, 2x + y + 10 = 0 Sol: The given equations are: Page 3 Exercise 3C 1. Solve the system of equations by using the method of cross multiplication: x + 2y + 1 = 0, 2x 3y 12 = 0. Sol: The given equations are: 2x 3y Here a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = -3 and c2 = -12 By cross multiplication, we have: = = = = = = x = = 3, y = = -2 Hence, x = 3 and y = -2 is the required solution. 2. Solve the system of equations by using the method of cross multiplication: 3x - 2y + 3 = 0, 4x + 3y 47 = 0 Sol: The given equations are: 3x - 2y 4x + 3y Here a1 = 3, b1 = -2, c1 = 3, a2 = 4, b2 = 3 and c2 = -47 By cross multiplication, we have: = = = = = = x = = 5, y = = 9 Hence, x = 5 and y = 9 is the required solution. 3. Solve the system of equations by using the method of cross multiplication: 6x - 5y - 16 = 0, 7x - 13y + 10 = 0 Sol: The given equations are: 6x - 5y - 7x - Here a1 = 6, b1 = -5, c1 = -16, a2 = 7, b2 = -13 and c2 = 10 By cross multiplication, we have: = = = = = = x = = 6, y = = 4 Hence, x = 6 and y = 4 is the required solution. 4. Solve the system of equations by using the method of cross multiplication: 3x + 2y + 25 = 0, 2x + y + 10 = 0 Sol: The given equations are: Here a1 = 3, b1 = 2, c1 = 25, a2 = 2, b2 = 1 and c2 = 10 By cross multiplication, we have: = = = = = = x = = 5, y = = -20 Hence, x = 5 and y = -20 is the required solution. 5. Solve the system of equations by using the method of cross multiplication: 2x + 5y 1 = 0, 2x + 3y 3 = 0 Sol: The given equations may be written as: 2x + 5y 2x + 3y Here a1 = 2, b1 = 5, c1 = -1, a2 = 2, b2 = 3 and c2 = -3 By cross multiplication, we have: = = = = = = x = = 3, y = = -1 Hence, x = 3 and y = -1 is the required solution. 6. Solve the system of equations by using the method of cross multiplication: 2x + y 35 = 0, 3x + 4y 65 = 0 Sol: The given equations may be written as: 2x + y Page 4 Exercise 3C 1. Solve the system of equations by using the method of cross multiplication: x + 2y + 1 = 0, 2x 3y 12 = 0. Sol: The given equations are: 2x 3y Here a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = -3 and c2 = -12 By cross multiplication, we have: = = = = = = x = = 3, y = = -2 Hence, x = 3 and y = -2 is the required solution. 2. Solve the system of equations by using the method of cross multiplication: 3x - 2y + 3 = 0, 4x + 3y 47 = 0 Sol: The given equations are: 3x - 2y 4x + 3y Here a1 = 3, b1 = -2, c1 = 3, a2 = 4, b2 = 3 and c2 = -47 By cross multiplication, we have: = = = = = = x = = 5, y = = 9 Hence, x = 5 and y = 9 is the required solution. 3. Solve the system of equations by using the method of cross multiplication: 6x - 5y - 16 = 0, 7x - 13y + 10 = 0 Sol: The given equations are: 6x - 5y - 7x - Here a1 = 6, b1 = -5, c1 = -16, a2 = 7, b2 = -13 and c2 = 10 By cross multiplication, we have: = = = = = = x = = 6, y = = 4 Hence, x = 6 and y = 4 is the required solution. 4. Solve the system of equations by using the method of cross multiplication: 3x + 2y + 25 = 0, 2x + y + 10 = 0 Sol: The given equations are: Here a1 = 3, b1 = 2, c1 = 25, a2 = 2, b2 = 1 and c2 = 10 By cross multiplication, we have: = = = = = = x = = 5, y = = -20 Hence, x = 5 and y = -20 is the required solution. 5. Solve the system of equations by using the method of cross multiplication: 2x + 5y 1 = 0, 2x + 3y 3 = 0 Sol: The given equations may be written as: 2x + 5y 2x + 3y Here a1 = 2, b1 = 5, c1 = -1, a2 = 2, b2 = 3 and c2 = -3 By cross multiplication, we have: = = = = = = x = = 3, y = = -1 Hence, x = 3 and y = -1 is the required solution. 6. Solve the system of equations by using the method of cross multiplication: 2x + y 35 = 0, 3x + 4y 65 = 0 Sol: The given equations may be written as: 2x + y 3x + 4y Here a1 = 2, b1 = 1, c1 = -35, a2 = 3, b2 = 4 and c2 = -65 By cross multiplication, we have: = = = = = = x = = 15, y = = 5 Hence, x = 15 and y = 5 is the required solution. 7. Solve the system of equations by using the method of cross multiplication: 7x - 2y 3 = 0, 11x - y 8 = 0. Sol: The given equations may be written as: 7x - 2y 11x - y Here a1 = 7, b1 = -2, c1 = -3, a2 = 11, b2 = - and c2 = -8 By cross multiplication, we have: = = = = = = x = = 1, y = = 2 Hence, x = 1 and y = 2 is the required solution. 8. Solve the system of equations by using the method of cross multiplication: + 4 = 0, - = 0 Page 5 Exercise 3C 1. Solve the system of equations by using the method of cross multiplication: x + 2y + 1 = 0, 2x 3y 12 = 0. Sol: The given equations are: 2x 3y Here a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = -3 and c2 = -12 By cross multiplication, we have: = = = = = = x = = 3, y = = -2 Hence, x = 3 and y = -2 is the required solution. 2. Solve the system of equations by using the method of cross multiplication: 3x - 2y + 3 = 0, 4x + 3y 47 = 0 Sol: The given equations are: 3x - 2y 4x + 3y Here a1 = 3, b1 = -2, c1 = 3, a2 = 4, b2 = 3 and c2 = -47 By cross multiplication, we have: = = = = = = x = = 5, y = = 9 Hence, x = 5 and y = 9 is the required solution. 3. Solve the system of equations by using the method of cross multiplication: 6x - 5y - 16 = 0, 7x - 13y + 10 = 0 Sol: The given equations are: 6x - 5y - 7x - Here a1 = 6, b1 = -5, c1 = -16, a2 = 7, b2 = -13 and c2 = 10 By cross multiplication, we have: = = = = = = x = = 6, y = = 4 Hence, x = 6 and y = 4 is the required solution. 4. Solve the system of equations by using the method of cross multiplication: 3x + 2y + 25 = 0, 2x + y + 10 = 0 Sol: The given equations are: Here a1 = 3, b1 = 2, c1 = 25, a2 = 2, b2 = 1 and c2 = 10 By cross multiplication, we have: = = = = = = x = = 5, y = = -20 Hence, x = 5 and y = -20 is the required solution. 5. Solve the system of equations by using the method of cross multiplication: 2x + 5y 1 = 0, 2x + 3y 3 = 0 Sol: The given equations may be written as: 2x + 5y 2x + 3y Here a1 = 2, b1 = 5, c1 = -1, a2 = 2, b2 = 3 and c2 = -3 By cross multiplication, we have: = = = = = = x = = 3, y = = -1 Hence, x = 3 and y = -1 is the required solution. 6. Solve the system of equations by using the method of cross multiplication: 2x + y 35 = 0, 3x + 4y 65 = 0 Sol: The given equations may be written as: 2x + y 3x + 4y Here a1 = 2, b1 = 1, c1 = -35, a2 = 3, b2 = 4 and c2 = -65 By cross multiplication, we have: = = = = = = x = = 15, y = = 5 Hence, x = 15 and y = 5 is the required solution. 7. Solve the system of equations by using the method of cross multiplication: 7x - 2y 3 = 0, 11x - y 8 = 0. Sol: The given equations may be written as: 7x - 2y 11x - y Here a1 = 7, b1 = -2, c1 = -3, a2 = 11, b2 = - and c2 = -8 By cross multiplication, we have: = = = = = = x = = 1, y = = 2 Hence, x = 1 and y = 2 is the required solution. 8. Solve the system of equations by using the method of cross multiplication: + 4 = 0, - = 0 Sol: The given equations may be written as: + - Here a1 = , b1 = , c1 = -4, a2 = , b2 = - and c2 = - By cross multiplication, we have: = = = = = = x = = 18, y = = 15 Hence, x = 18 and y = 15 is the required solution. 9. Solve the system of equations by using the method of cross multiplication: + = 7, + = 17 Sol: Taking = u and = v, the given equations become: u + v = 7 2u + 3v = 17 The given equations may be written as: u + v 2u + 3v Here, a1 = 1, b1 = 1, c1 = -7, a2 = 2, b2 = 3 and c2 = -17 By cross multiplication, we have: = = = = = = u = = 4, v = = 3 = 4, = 3 x = , y =Read More

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