Page 1 Exercise 3D 1. Show that the following system of equations has a unique solution: 3x + 5y = 12, 5x + 3y = 4. Also, find the solution of the given system of equations. Sol: The given system of equations is: Page 2 Exercise 3D 1. Show that the following system of equations has a unique solution: 3x + 5y = 12, 5x + 3y = 4. Also, find the solution of the given system of equations. Sol: The given system of equations is: 3x + 5y = 12 5x + 3y = 4 These equations are of the forms: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 3, b1= 5, c1 = -12 and a2 = 5, b2 = 3, c2 = -4 For a unique solution, we must have: , i.e., Hence, the given system of equations has a unique solution. Again, the given equations are: On multiplying (i) by 3 and (ii) by 5, we get: 9x + 15y = 36 On subtracting (iii) from (iv), we get: 16x = -16 x = -1 On substituting x = -1 in (i), we get: 3(-1) + 5y = 12 5y = (12 + 3) = 15 y = 3 Hence, x = -1 and y = 3 is the required solution. 2. Show that the following system of equations has a unique solution: 2x - 3y = 17, 4x + y = 13. Also, find the solution of the given system of equations. Sol: The given system of equations is: 2x - 3y - 4x + y - The given equations are of the form a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= -3, c1 = -17 and a2 = 4, b2 = 1, c2 = -13 Now, = = and = = -3 Since, , therefore the system of equations has unique solution. Using cross multiplication method, we have Page 3 Exercise 3D 1. Show that the following system of equations has a unique solution: 3x + 5y = 12, 5x + 3y = 4. Also, find the solution of the given system of equations. Sol: The given system of equations is: 3x + 5y = 12 5x + 3y = 4 These equations are of the forms: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 3, b1= 5, c1 = -12 and a2 = 5, b2 = 3, c2 = -4 For a unique solution, we must have: , i.e., Hence, the given system of equations has a unique solution. Again, the given equations are: On multiplying (i) by 3 and (ii) by 5, we get: 9x + 15y = 36 On subtracting (iii) from (iv), we get: 16x = -16 x = -1 On substituting x = -1 in (i), we get: 3(-1) + 5y = 12 5y = (12 + 3) = 15 y = 3 Hence, x = -1 and y = 3 is the required solution. 2. Show that the following system of equations has a unique solution: 2x - 3y = 17, 4x + y = 13. Also, find the solution of the given system of equations. Sol: The given system of equations is: 2x - 3y - 4x + y - The given equations are of the form a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= -3, c1 = -17 and a2 = 4, b2 = 1, c2 = -13 Now, = = and = = -3 Since, , therefore the system of equations has unique solution. Using cross multiplication method, we have = = = = = = = = x = , y = x = 4, y = -3 Hence, x = 4 and y = -3. 3. Show that the following system of equations has a unique solution: + = 3, x 2y = 2. Also, find the solution of the given system of equations. Sol: The given system of equations is: + = 3 = 3 2x + 3y = 18 2x + 3y 1 and x 2y = 2 x 2y These equations are of the forms: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= 3, c1 = -18 and a2 = 1, b2 = -2, c2 = -2 For a unique solution, we must have: , i.e., Hence, the given system of equations has a unique solution. Again, the given equations are: 2x + 3y x 2y On multiplying (i) by 2 and (ii) by 3, we get: 4x + 6y 3x - 6y On adding (v) from (vi), we get: 7x = 42 x = 6 On substituting x = 6 in (iii), we get: Page 4 Exercise 3D 1. Show that the following system of equations has a unique solution: 3x + 5y = 12, 5x + 3y = 4. Also, find the solution of the given system of equations. Sol: The given system of equations is: 3x + 5y = 12 5x + 3y = 4 These equations are of the forms: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 3, b1= 5, c1 = -12 and a2 = 5, b2 = 3, c2 = -4 For a unique solution, we must have: , i.e., Hence, the given system of equations has a unique solution. Again, the given equations are: On multiplying (i) by 3 and (ii) by 5, we get: 9x + 15y = 36 On subtracting (iii) from (iv), we get: 16x = -16 x = -1 On substituting x = -1 in (i), we get: 3(-1) + 5y = 12 5y = (12 + 3) = 15 y = 3 Hence, x = -1 and y = 3 is the required solution. 2. Show that the following system of equations has a unique solution: 2x - 3y = 17, 4x + y = 13. Also, find the solution of the given system of equations. Sol: The given system of equations is: 2x - 3y - 4x + y - The given equations are of the form a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= -3, c1 = -17 and a2 = 4, b2 = 1, c2 = -13 Now, = = and = = -3 Since, , therefore the system of equations has unique solution. Using cross multiplication method, we have = = = = = = = = x = , y = x = 4, y = -3 Hence, x = 4 and y = -3. 3. Show that the following system of equations has a unique solution: + = 3, x 2y = 2. Also, find the solution of the given system of equations. Sol: The given system of equations is: + = 3 = 3 2x + 3y = 18 2x + 3y 1 and x 2y = 2 x 2y These equations are of the forms: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= 3, c1 = -18 and a2 = 1, b2 = -2, c2 = -2 For a unique solution, we must have: , i.e., Hence, the given system of equations has a unique solution. Again, the given equations are: 2x + 3y x 2y On multiplying (i) by 2 and (ii) by 3, we get: 4x + 6y 3x - 6y On adding (v) from (vi), we get: 7x = 42 x = 6 On substituting x = 6 in (iii), we get: 2(6) + 3y = 18 3y = (18 - 12) = 6 y = 2 Hence, x = 6 and y = 2 is the required solution. 4. Find the value of k for which the system of equations has a unique solution: 2x + 3y = 5, kx - 6y = 8. Sol: The given system of equations are 2x + 3y 5 = 0 kx - 6y - 8 = 0 This system is of the form: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= 3, c1 = -5 and a2 = k, b2 = -6, c2 = -8 Now, for the given system of equations to have a unique solution, we must have: -4 Hence, -4 5. Find the value of k for which the system of equations has a unique solution: x ky = 2, 3x + 2y + 5=0. Sol: The given system of equations are x - ky 2 = 0 3x + 2y + 5 = 0 This system of equations is of the form: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 1, b1= -k, c1 = -2 and a2 = 3, b2 = 2, c2 = 5 Now, for the given system of equations to have a unique solution, we must have: - Hence, - . Page 5 Exercise 3D 1. Show that the following system of equations has a unique solution: 3x + 5y = 12, 5x + 3y = 4. Also, find the solution of the given system of equations. Sol: The given system of equations is: 3x + 5y = 12 5x + 3y = 4 These equations are of the forms: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 3, b1= 5, c1 = -12 and a2 = 5, b2 = 3, c2 = -4 For a unique solution, we must have: , i.e., Hence, the given system of equations has a unique solution. Again, the given equations are: On multiplying (i) by 3 and (ii) by 5, we get: 9x + 15y = 36 On subtracting (iii) from (iv), we get: 16x = -16 x = -1 On substituting x = -1 in (i), we get: 3(-1) + 5y = 12 5y = (12 + 3) = 15 y = 3 Hence, x = -1 and y = 3 is the required solution. 2. Show that the following system of equations has a unique solution: 2x - 3y = 17, 4x + y = 13. Also, find the solution of the given system of equations. Sol: The given system of equations is: 2x - 3y - 4x + y - The given equations are of the form a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= -3, c1 = -17 and a2 = 4, b2 = 1, c2 = -13 Now, = = and = = -3 Since, , therefore the system of equations has unique solution. Using cross multiplication method, we have = = = = = = = = x = , y = x = 4, y = -3 Hence, x = 4 and y = -3. 3. Show that the following system of equations has a unique solution: + = 3, x 2y = 2. Also, find the solution of the given system of equations. Sol: The given system of equations is: + = 3 = 3 2x + 3y = 18 2x + 3y 1 and x 2y = 2 x 2y These equations are of the forms: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= 3, c1 = -18 and a2 = 1, b2 = -2, c2 = -2 For a unique solution, we must have: , i.e., Hence, the given system of equations has a unique solution. Again, the given equations are: 2x + 3y x 2y On multiplying (i) by 2 and (ii) by 3, we get: 4x + 6y 3x - 6y On adding (v) from (vi), we get: 7x = 42 x = 6 On substituting x = 6 in (iii), we get: 2(6) + 3y = 18 3y = (18 - 12) = 6 y = 2 Hence, x = 6 and y = 2 is the required solution. 4. Find the value of k for which the system of equations has a unique solution: 2x + 3y = 5, kx - 6y = 8. Sol: The given system of equations are 2x + 3y 5 = 0 kx - 6y - 8 = 0 This system is of the form: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= 3, c1 = -5 and a2 = k, b2 = -6, c2 = -8 Now, for the given system of equations to have a unique solution, we must have: -4 Hence, -4 5. Find the value of k for which the system of equations has a unique solution: x ky = 2, 3x + 2y + 5=0. Sol: The given system of equations are x - ky 2 = 0 3x + 2y + 5 = 0 This system of equations is of the form: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 1, b1= -k, c1 = -2 and a2 = 3, b2 = 2, c2 = 5 Now, for the given system of equations to have a unique solution, we must have: - Hence, - . 6. Find the value of k for which the system of equations has a unique solution: 5x 7y = 5, 2x + ky = 1. Sol: The given system of equations are 5x - 7y 2x + ky - This system is of the form: a1x+b1y+c1 = 0 a2x+b2y+c2 = 0 where, a1 = 5, b1= -7, c1 = -5 and a2 = 2, b2 = k, c2 = -1 Now, for the given system of equations to have a unique solution, we must have: - Hence, - . 7. Find the value of k for which the system of equations has a unique solution: 4x + ky + 8=0, x + y + 1 = 0. Sol: The given system of equations are 4x + ky + 8 = 0 x + y + 1 = 0 This system is of the form: a1x+b1y+c1 = 0 a2x+b2y+c2 = 0 where, a1 = 4, b1= k, c1 = 8 and a2 = 1, b2 = 1, c2 = 1 For the given system of equations to have a unique solution, we must have: Hence, 8. Find the value of k for which the system of equations has a unique solution: 4x - 5y = k, 2x - 3y = 12. Sol:Read More

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