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Chapter 3: Pair of Linear Equations in Two Variables Exercise – 3.5 Question: 1 In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution, x − 3y − 3 = 0, x − 3y − 3 = 0 3x − 9y − 2 = 0, 3x − 9y − 2 = 0 Solution: The given system may be written as x − 3y − 3 = 0 3x − 9y − 2 = 0 The given system of equation is of the form a_{2}x + b_{2}y − c_{2} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 1, b_{1} = −3, c_{1} = −3 a_{2} = 3, b_{2} = −9, c_{2} = −2 We have, Therefore, the given equation has no solution. Question: 2 In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution, 2x + y − 5 = 0 4x + 2y − 10 = 0 Solution: The given system may be written as 2x + y − 5 = 0 4x + 2y − 10 = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 2, b_{1} = 1, c_{1} = −5 a_{2} = 4, b_{2} = 2, c_{2} = −10 We have, Therefore, the given equation has infinitely many solution. Question: 3 In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution, 3x − 5y = 20 6x − 10y = 40 Solution: The given system may be written as 3x − 5y = 20 6x − 10y = 40 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 3, b_{1} = −5, c_{1} = − 20 a_{2} = 6, b_{2} = −10, c_{2 }= − 40 We have, Therefore, the given equation has infinitely many solution. Question: 4 In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution, x − 2y − 8 = 0 5x − 10y − 10 = 0 Solution: The given system may be written as x − 2y − 8 = 0 5x − 10y − 10 = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 1, b_{1} = −2, c_{1} = −8 a_{2} = 5, b_{2} = −10, c_{2} = −10 We have, Therefore, the given equation has no solution. Question: 5 Find the value of k for each of the following system of equations which have a unique solution kx + 2y − 5 = 0 3x + y − 1 = 0 Solution: The given system may be written as kx + 2y − 5 = 0 3x + y − 1 = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1 }= k, b_{1 }= 2, c_{1} = −5 a_{2} = 3, b_{2} = 1, c_{2} = −1 For unique solution, we have Therefore, the given system will have unique solution for all real values of k other than 6. Question: 6 Find the value of k for each of the following system of equations which have a unique solution 4x + ky + 8 = 0 2x + 2y + 2 = 0 Solution: The given system may be written as 4x + ky + 8 = 0 2x + 2y + 2 = 0 The given system of equation is of the form a_{1}x +b_{1}y − c_{1 }= 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 4, b_{1} = k, c_{1} = 8 a_{2 }= 2, b_{2} = 2, c_{2} = 2 For unique solution, we have Therefore, the given system will have unique solution for all real values of k other than 4. Question: 7 Find the value of k for each of the following system of equations which have a unique solution 4x − 5y = k 2x − 3y = 12 Solution: The given system may be written as 4x − 5y − k = 0 2x − 3y − 12 = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1 }= 4, b_{1} = −5, c_{1} = −k a_{2} = 2, b_{2 }= -3, c_{2 }= -12 For unique solution, we have ⇒ k can have any real values. Therefore, the given system will have unique solution for all real values of k. Question: 8 Find the value of k for each of the following system of equations which have a unique solution x + 2y = 3 5x + ky + 7 = 0 Solution: The given system may be written as x + 2y = 3 5x + ky + 7 = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where a_{1 }= 1, b_{1} = 2, c_{1} = −3 a_{2 }= 5, b_{2} = k, c_{2} = 7 For unique solution, we have Therefore, the given system will have unique solution for all real values of k other than 10. Question: 9 Find the value of k for which each of the following system of equations having infinitely many solution: 2x + 3y − 5 = 0 6x − ky − 15 = 0 Solution: The given system may be written as 2x + 3y − 5 = 0 6x − ky − 15 = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 2, b_{1} = 3, c_{1} = −5 a_{2} = 6, b_{2} = k, c_{2} = −15 For unique solution, we have Therefore, the given system of equation will have infinitely many solutions, if k = 9. Question: 10 Find the value of k for which each of the following system of equations having infinitely many solution: 4x + 5y = 3 x + 15y = 9 Solution: The given system may be written as 4x + 5y = 3 kx +15y = 9 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 4, b_{1} = 5, c_{1} = 3 a_{2} = k, b_{2 }= 15, c_{2} = 9 For unique solution, we have Therefore, the given system will have infinitely many solutions if k = 12. Question: 11 Find the value of k for which each of the following system of equations having infinitely many solution: kx − 2y + 6 = 0 4x + 3y + 9 = 0 Solution: The given system may be written as kx − 2y + 6 = 0 4x + 3y + 9 = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = k, b_{1} = −2, c_{1} = 6 a_{2} = 4, b_{2} = −3, c_{2} = 9 For unique solution, we have Therefore, the given system of equations will have infinitely many solutions, if k = 8/3. Question: 12 Find the value of k for which each of the following system of equations having infinitely many solution: 8x + 5y = 9 kx + 10y = 19 Solution: The given system may be written as 8x + 5y = 9 kx + 10y = 19 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2 }=0 Where, a_{1} = 8, b_{1} = 5, c_{1 }= −9 a_{2} = k, b_{2} = 10, c_{2} = −18 a_{2} = k, b_{2} = 10,c_{2 }= −18 For unique solution, we have Therefore, the given system of equations will have infinitely many solutions, if k = 16. Question: 13 Find the value of k for which each of the following system of equations having infinitely many solution: 2x − 3y = 7 (k + 2)x − (2k + 1)y = 3(2k − 1) Solution: The given system may be written as 2x − 3y = 7 (k + 2)x − (2k + 1)y = 3(2k − 1) The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 2, b_{1} = −3, c_{1} = −7 a_{2 }= k, b_{2} = − (2k + 1), c_{2} = −3(2k − 1) For unique solution, we have ⇒ 2(2k + 1) = 3(k + 2) and 3 × 3(2k − 1) = 7(2k + 1) ⇒ 4k + 2 = 3k + 6 and 18k − 9 = 14k + 7 ⇒ k = 4 and 4k = 16 ⇒ k = 4 Therefore, the given system of equations will have infinitely many solutions, if k = 4. Question: 14 Find the value of k for which each of the following system of equations having infinitely many solution: 2x + 3y = 2 (k + 2)x + (2k + 1)y = 2(k − 1) Solution: The given system may be written as 2x + 3y = 2 (k + 2)x + (2k + 1)y = 2(k − 1) The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 2, b_{1} = 3,c_{1} = −2 a_{2} = (k + 2), b_{2} = (2k + 1),c_{2} = −2(k − 1) For unique solution, we have ⇒ 2(2k + 1) = 3(k + 2) and 3(k − 1) = (2k + 1) ⇒ 4k + 2 = 3k + 6 and 3k − 3 = 2k + 1 ⇒ k = 4 and k = 4 Therefore, the given system of equations will have infinitely many solutions, if k = 4. Question: 15 Find the value of k for which each of the following system of equations having infinitely many solution: x + (k + 1)y = 4 (k + 1)x + 9y = (5k + 2) Solution: The given system may be written as x + (k + 1)y = 4 (k + 1)x + 9y = (5k + 2) The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 1, b_{1} = (k + 1), c_{1} = −4 a_{2} = (k + 1), b_{2} = 9, c_{2} = − (5k + 2) For unique solution, we have ⇒ 9 = (k + 1)^{2} and (k + 1)(5k + 2) = 36 ⇒ 9 = k^{2} + 2k + 1 and 5k^{2} + 2k + 5k + 2 = 36 ⇒ k^{2 }+ 2k − 8 = 0 and 5k^{2} + 7k − 34 = 0 ⇒ k^{2} + 4k − 2k − 8 = 0 and 5k^{2} + 17k − 10k − 34 = 0 ⇒ k(k + 4) −2 (k + 4) = 0 and (5k + 17) − 2 (5k + 17) = 0 ⇒ (k + 4)(k − 2) = 0 and (5k + 17)(k − 2) = 0 ⇒ k = - 4 or k = 2 and k = -17/5 or k = 2 Thus, k = 2 satisfies both the condition. Therefore, the given system of equations will have infinitely many solutions, if k = 2. Question: 16 Find the value of k for which each of the following system of equations having infinitely many solution: kx + 3y = 2k + 1 2(k + 1)x + 9y = (7k + 1) Solution: The given system may be written as kx + 3y = 2k + 1 2(k + 1)x + 9y = (7k + 1) The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1 }= k, b_{1} = 3, c_{1} = −(2k + 1) a_{2} = 2(k + 1), b_{2} = 9, c_{2} = −(7k + 1) For unique solution, we have ⇒ 9k = 3 × 2(k + 1) and 3(7k + 1) = 9(2k + 1) ⇒ 9k − 6k = 6 and 21k − 18k = 9 − 3 ⇒ 3k = 6 ⇒ k = 2 and k = 2 Therefore, the given system of equations will have infinitely many solutions, if k = 2. Question: 17 Find the value of k for which each of the following system of equations having infinitely many solution: 2x + (k − 2)y = k 6x + (2k − 1)y = (2k + 5) Solution: The given system may be written as 2x +( k − 2)y = k 6x + (2k − 1)y = (2k + 5) The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 2,b_{1} = (k − 2),c_{1} = −k a_{2} = 6,b_{2} = (2k − 1),c_{2} = −(2k + 5) For unique solution, we have ⇒ 2k − 3k = −6 + 1 and k + k = 10 ⇒ −k = −5 and 2k = 10 ⇒ k = 5 and k = 5 Therefore, the given system of equations will have infinitely many solutions, if k = 5. Question: 18 Find the value of k for which each of the following system of equations having infinitely many solution: 2x + 3y = 72x + 3y = 7 (k + 1)x + (2k − 1)y = (4k + 1) Solution: The given system may be written as 2x + 3y = 7 (k + 1)x + (2k − 1)y = (4k + 1) The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2 }= 0 Where, a_{1} = 2, b_{1} = 3, c_{1} = −7 a_{2 }= k + 1, b_{2} = 2k − 1, c_{2} = −(4k + 1) For unique solution, we have Extra close brace or missing open brace ⇒ 4k − 2 = 3k + 3 and 12k + 3 = 14k − 7 ⇒ k = 5 and 2k = 10 ⇒ k = 5 and k = 5 Therefore, the given system of equations will have infinitely many solutions, if k = 5. Question: 19 Find the value of k for which each of the following system of equations having infinitely many solution: 2x + 3y = k (k − 1)x + (k + 2)y = 3k Solution: The given system may be written as 2x + 3y = k (k − 1)x + (k + 2)y = 3k The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1 }= 2,b_{1 }= 3, c_{1} = −k a_{2} = k − 1, b_{2 }= k + 2, c_{2} = −3k For unique solution, we have Extra close brace or missing open brace ⇒ 2k + 4 = 3k − 3 and 9 = k + 2 ⇒ 2k + 4 = 3k − 3 and 9 = k + 2 ⇒ k = 7 and k = 7 Therefore, the given system of equations will have infinitely many solutions, if k = 7. Question: 20 Find the value of k for which the following system of equation has no solution: kx − 5y = 2 6x + 2y = 7 Solution: The given system may be written as kx − 5y = 2 6x + 2y = 7 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1 }= k, b_{1} = −5, c_{1 }= −2 a_{2} = 6 b_{2 }= 2, c_{2} = −7 For no solution, we have ⇒ 2k = -30 ⇒ k = -15 Therefore, the given system of equations will have no solutions, if k = −15. Question: 21 Find the value of k for which the following system of equation has no solution: x + 2y = 0 2x + ky = 5 Solution: The given system may be written as x_{2}y = 0 2x + ky = 5 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 1, b_{1} = 2, c_{1} = 0 a_{2} = 2, b_{2} = k, c_{2} = −5 For no solution, we have ⇒ k = 4 Therefore, the given system of equations will have no solutions, if k = 4. Question: 22 Find the value of k for which the following system of equation has no solution: 3x − 4y + 7 = 0 kx + 3y − 5 = 0 Solution: The given system may be written as 3x − 4y + 7 = 0 kx + 3y − 5 = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 3, b_{1} = −4, c_{1} = 7 a_{2} = k, b_{2} = 3, c_{2} = −5 For no solution, we have Therefore, the given system of equations will have no solutions, if k = - 9 /4. Question: 23 Find the value of k for which the following system of equation has no solution: 2x − ky + 3 = 0 3x + 2y − 1 = 0 Solution: The given system may be written as 2x − ky + 3 = 0 3x + 2y − 1 = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 2, b_{1} = −k, c_{1} = 3 a_{2} = 3, b_{2} = 2, c_{2} = −1 For no solution, we have Therefore, the given system of equations will have no solutions, if k = – 4/3. Question: 24 Find the value of k for which the following system of equation has no solution: 2x + ky − 11 = 0 5x − 7y − 5 = 0 Solution: The given system may be written as 2x + ky − 11 = 0 5x − 7y − 5 = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 2, b_{1} = k, c_{1} = −11 a_{2} = 5, b_{2} = −7, c_{2} = −5a_{2} = 5, b_{2 }= −7, c_{2} = − 5 For no solution, we have Therefore, the given system of equations will have no solutions, if k = -14/5. Question: 25 Find the value of k for which the following system of equation has no solution: kx + 3y = 3 12x + ky = 6 Solution: The given system may be written as kx + 3y = 3 12x + ky = 6 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = k, b_{1} = 3, c_{1} = −3 a_{2 }= 12, b_{2} = k, c_{2} = − 6 For no solution, we have ⇒ k^{2} = 36 ⇒ k = + 6 or −6 From (i) Therefore, the given system of equations will have no solutions, if k = − 6. Question: 26 For what value of a, the following system of equation will be inconsistent? 4x + 6y − 11 = 0 2x + ay − 7 = 0 Solution: The given system may be written as 4x + 6y − 11 = 0 2x + ay − 7 = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 4, b_{1} = 6, c_{1} = −11 a_{2} = 2, b_{2} = a, c_{2} = −7 For unique solution, we have Therefore, the given system of equations will be inconsistent, if a = 3. Question: 27 For what value of a, the following system of equation have no solution? ax + 3y = a − 3 12x + ay = a Solution: The given system may be written as ax + 3y = a − 3 12x + ay = a The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = a, b_{1 }= 3, c_{1} = - (a − 3) a_{2} = 12, b_{2} = a, c_{2 }= − a For unique solution, we have And, ⇒ a^{2 }= 36 ⇒ a = + 6 or – 6? a ≠ 6 ⇒ a = – 6 Therefore, the given system of equations will have no solution, if a = − 6. Question: 28 Find the value of a, for which the following system of equation have (i) Unique solution (ii) No solution kx + 2y = 5 3x + y = 1 Solution: The given system may be written as kx + 2y − 5 = 0 3x + y − 1 = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = k, b_{1} = 2, c_{1} = −5 a_{2} = 3, b_{2} = 1, c_{2} = −1 (i) For unique solution, we have Therefore, the given system of equations will have unique solution, if k ≠ 6 k ≠ 6. (ii) For no solution, we have Therefore, the given system of equations will have no solution, if a = 6. Question: 29 For what value of c, the following system of equation have infinitely many solution (where c ≠ 0 c ≠ 0)? 6x + 3y = c − 3 12x + cy = c Solution: The given system may be written as 6x + 3y − (c − 3) = 0 12x + cy − c = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 6, b_{1} = 3, c_{1} = −(c − 3) a_{2 }= 12, b_{2} = c, c_{2} = - c For infinitely many solution, we have ⇒ c = 6 and c – 3 = 3 ⇒ c = 6 and c = 6 Therefore, the given system of equations will have infinitely many solution, if c = 6. Question: 30 Find the value of k, for which the following system of equation have (i) Unique solution (ii) No solution (iii) Infinitely many solution 2x + ky = 1 3x − 5y = 7 Solution: The given system may be written as 2x + ky = 1 3x − 5y = 7 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 2, b_{1} = k, c_{1} = −1a_{1 }= 2, b_{1} = k, c_{1} = −1 a_{2 }= 3, b_{2} = −5, c_{2} = −7 (i) For unique solution, we have Therefore, the given system of equations will have unique solution, if k ≠ -10/3. (ii) For no solution, we have Therefore, the given system of equations will have no solution, if k = -10)/3. (iii) For the given system to have infinitely many solution, we have So there is no value of k for which the given system of equation has infinitely many solution. Question: 31 For what value of k, the following system of equation will represent the coincident lines? x + 2y + 7 = 0 2x + ky + 14 = 0 Solution: The given system may be written as x + 2y + 7 = 0 2x + ky + 14 = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 1, b_{1} = 2, c_{1} = 7 a_{2} = 2, b_{2} = k, c_{2} = 14 The given system of equation will represent the coincident lines if they have infinitely many solution. Therefore, the given system of equations will have infinitely many solution, if k = 4. Question: 32 Find the value of k, for which the following system of equation have unique solution. ax + by = c lx + my = n Solution: The given system may be written as ax + by − c = 0 lx + my − n = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2 }= 0 Where, a_{1} = a, b_{1} = b, c_{1 }= − c a_{2 }= l, b_{2} = m, c_{2} = − n For unique solution, we have Therefore, the given system of equations will have unique solution, if am ≠ bl. Question: 33 Find the value of a and b such that the following system of linear equation have infinitely many solution: (2a − 1)x + 3y − 5 = 0 3x + (b − 1)y − 2 = 0 Solution: The given system of equation may be written as, (2a − 1)x + 3y − 5 = 0 3x + (b − 1)y − 2 = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = (2a − 1), b_{1} = 3, c_{1} = −5 a_{2} = 3, b_{2} = b − 1, c_{2} = −2 The given system of equation will have infinitely many solution, if ⇒ 2(2a − 1) = 15 and 6 = 5(b − 1) ⇒ 4a − 2 = 15 and 6 = 5b − 5 ⇒ 4a = 17 and 5b = 11 Question: 34 Find the value of a and b such that the following system of linear equation have infinitely many solution: 2x − 3y = 7 (a + b)x − (a + b − 3)y = 4a + b Solution: The given system of equation may be written as, 2x − 3y − 7 = 0 (a + b)x − (a + b − 3)y − (4a + b) = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 2, b_{1} = −3, c_{1} = −7 a_{2} = (a + b), b_{2} = −(a + b − 3), c_{2 }= −(4a + b) The given system of equation will have infinitely many solution, if ⇒ 2(a + b − 3) = 3(a + b) and 3(4a + b) = 7(a + b − 3) ⇒ 2a + 2b − 6 = 3a + 3b and 12a + 3b = 7a + 7b − 21 ⇒ a + b = −6 and 5a − 4b = −21 ⇒ a = − 6 − b Substituting the value of a in 5a − 4b = −21 we have 5( – b – 6) – 4b = – 21 ⇒ − 5b − 30 − 4b = − 21 ⇒ 9b = − 9 ⇒ b = −1 As a = – 6 – b ⇒ a = − 6 + 1 = − 5 Hence the given system of equation will have infinitely many solution if a = – 5 and b = –1. Question: 35 Find the value of p and q such that the following system of linear equation have infinitely many solution: 2x − 3y = 9 (p + q)x + (2p − q)y = 3(p + q + 1) Solution: The given system of equation may be written as, 2x − 3y − 9 = 0 (p + q)x + (2p − q)y − 3(p + q + 1) = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} =2, b_{1} = 3, c_{1 }= −9 a_{2} = (p + q), b_{2} = (2p − q), c_{2} = -3(p + q + 1) The given system of equation will have infinitely many solution, if 2(2p - q) = 3(p + q) and (p + q + 1) = 2p - q ⇒ 4p - 2q = 3p + 3q and -p + 2q = -1 ⇒ p = 5q and p - 2q = 1 Substituting the value of p in p - 2q = 1, we have 3q = 1 ⇒ q = 1/3 Substituting the value of p in p = 5q we have p = 5/3 Hence the given system of equation will have infinitely many solution if p = 5/3 and q = 1/3. Question: 36 Find the values of a and b for which the following system of equation has infinitely many solution: (i) (2a − 1)x + 3y = 5 3x + (b − 2)y = 3 (ii) 2x − (2a + 5)y = 5 (2b + 1)x − 9y = 15 (iii) (a − 1)x + 3y = 2 6x + (1 − 2b)y = 6 (iv) 3x + 4y = 12 (a + b)x + 2(a − b)y = 5a – 1 (v) 2x + 3y = 7 (a − 1)x + (a + 1)y = 3a − 1 (vi) 2x + 3y = 7 (a − 1)x + (a + 2)y = 3a Solution: (i) The given system of equation may be written as, (2a − 1)x + 3y − 5 = 0 3x + (b − 2)y − 3 = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 2a − 1, b_{1 }= 3, c_{1} = −5 a_{2} = 3, b_{2} = b − 2, c_{2 }= -3(p + q + 1) The given system of equation will have infinitely many solution, if 2a - 1 = 5 and - 9 = 5(b - 2) ⇒ a = 3 and -9 = 5b - 10 a = 3 and b = 1/5 Hence the given system of equation will have infinitely many solution if a = 3 and b = 1/5. (ii) The given system of equation may be written as, 2x − (2a + 5)y = 5 (2b + 1)x − 9y = 15 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 2, b_{1} = - (2a + 5), c_{1} = −5 a_{2} = (2b + 1), b_{2} = −9, c_{2} = −15 The given system of equation will have infinitely many solution, if Hence the given system of equation will have infinitely many solution if a = - 1 and b = 5/2. (iii) The given system of equation may be written as, (a − 1)x + 3y = 2 6x + (1 − 2b)y = 6 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = a-1, b_{1} = 3, c_{1} = −2 a_{2} = 6, b_{2} = 1 − 2b, c_{2} = −6 The given system of equation will have infinitely many solution, if ⇒ a – 1 = 2 and 1 – 2b = 9 ⇒ a - 1 = 2 and 1 - 2b = 9 ⇒ a = 3 and b = -4 ⇒ a = 3 and b = -4 Hence the given system of equation will have infinitely many solution if a = 3 and b = −4. (iv) The given system of equation may be written as, 3x + 4y − 12 = 0 (a + b)x + 2(a − b)y − (5a − 1) = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 3, b_{1} = 4, c_{1} = −12 a_{2 }= (a + b), b_{2} = 2(a − b), c_{2} = – (5a − 1) The given system of equation will have infinitely many solution, if ⇒ 3(a - b) = 2a + 2b and 2(5a - 1) = 12(a - b) ⇒ a = 5b and -2a = -12b + 2 Substituting a = 5b in -2a = -12b + 2, we have -2(5b) = -12b + 2 ⇒ −10b = −12b + 2 ⇒ b = 1 Thus a = 5 Hence the given system of equation will have infinitely many solution if a = 5 and b = 1. (v) The given system of equation may be written as, 2x + 3y − 7 = 0 (a − 1)x + (a + 1)y − (3a − 1) = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1} = 2, b_{1} = 3, c_{1} = −7 a_{2} = (a − 1), b_{2} = (a + 1), c_{2} = - (3a − 1) The given system of equation will have infinitely many solution, if ⇒ 2(a + 1) = 3(a - 1) and 3(3a - 1) = 7(a + 1) ⇒ 2a - 3a = -3 - 2 and 9a - 3 = 7a + 7 ⇒ a = 5 and a = 5 Hence the given system of equation will have infinitely many solution if a = 5 and b = 1. (vi) The given system of equation may be written as, 2x + 3y − 7 = 0 (a − 1)x + (a + 2)y − 3a = 0 The given system of equation is of the form a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0 Where, a_{1 }= 2, b_{1} = 3, c_{1} = −7 a_{2} = (a − 1), b_{2 }= (a + 2), c_{2} = −3a The given system of equation will have infinitely many solution, if ⇒ 2(a + 2) = 3(a - 1) and 3(3a) = 7(a + 2) ⇒ 2a + 4 = 3a - 3 and 9a = 7a + 14 ⇒ a = 7 and a = 7 Hence the given system of equation will have infinitely many solution if a = 7and b = 1.
In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution,
x − 3y − 3 = 0, x − 3y − 3 = 0
3x − 9y − 2 = 0, 3x − 9y − 2 = 0
The given system may be written as
x − 3y − 3 = 0
3x − 9y − 2 = 0
The given system of equation is of the form
a_{2}x + b_{2}y − c_{2} = 0 a_{2}x + b_{2}y − c_{2} = 0
Where, a_{1} = 1, b_{1} = −3, c_{1} = −3
a_{2} = 3, b_{2} = −9, c_{2} = −2
We have,
Therefore, the given equation has no solution.
2x + y − 5 = 0
4x + 2y − 10 = 0
2x + y − 5 = 0 4x + 2y − 10 = 0
a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0
Where, a_{1} = 2, b_{1} = 1, c_{1} = −5
a_{2} = 4, b_{2} = 2, c_{2} = −10
Therefore, the given equation has infinitely many solution.
3x − 5y = 20
6x − 10y = 40
3x − 5y = 20 6x − 10y = 40
Where, a_{1} = 3, b_{1} = −5, c_{1} = − 20
a_{2} = 6, b_{2} = −10, c_{2 }= − 40
x − 2y − 8 = 0
5x − 10y − 10 = 0
x − 2y − 8 = 0 5x − 10y − 10 = 0
Where, a_{1} = 1, b_{1} = −2, c_{1} = −8
a_{2} = 5, b_{2} = −10, c_{2} = −10
Find the value of k for each of the following system of equations which have a unique solution
kx + 2y − 5 = 0
3x + y − 1 = 0
a_{1}x + b_{1}y − c_{1} = 0
a_{2}x + b_{2}y − c_{2} = 0
Where, a_{1 }= k, b_{1 }= 2, c_{1} = −5
a_{2} = 3, b_{2} = 1, c_{2} = −1
For unique solution, we have
Therefore, the given system will have unique solution for all real values of k other than 6.
4x + ky + 8 = 0
2x + 2y + 2 = 0
4x + ky + 8 = 0 2x + 2y + 2 = 0
a_{1}x +b_{1}y − c_{1 }= 0 a_{2}x + b_{2}y − c_{2} = 0
Where, a_{1} = 4, b_{1} = k, c_{1} = 8
a_{2 }= 2, b_{2} = 2, c_{2} = 2
Therefore, the given system will have unique solution for all real values of k other than 4.
4x − 5y = k
2x − 3y = 12
4x − 5y − k = 0
2x − 3y − 12 = 0
Where, a_{1 }= 4, b_{1} = −5, c_{1} = −k
a_{2} = 2, b_{2 }= -3, c_{2 }= -12
⇒ k can have any real values.
Therefore, the given system will have unique solution for all real values of k.
x + 2y = 3
5x + ky + 7 = 0
Where a_{1 }= 1, b_{1} = 2, c_{1} = −3
a_{2 }= 5, b_{2} = k, c_{2} = 7
Therefore, the given system will have unique solution for all real values of k other than 10.
Find the value of k for which each of the following system of equations having infinitely many solution:
2x + 3y − 5 = 0
6x − ky − 15 = 0
2x + 3y − 5 = 0 6x − ky − 15 = 0
Where, a_{1} = 2, b_{1} = 3, c_{1} = −5
a_{2} = 6, b_{2} = k, c_{2} = −15
Therefore, the given system of equation will have infinitely many solutions, if k = 9.
4x + 5y = 3
x + 15y = 9
kx +15y = 9
Where, a_{1} = 4, b_{1} = 5, c_{1} = 3
a_{2} = k, b_{2 }= 15, c_{2} = 9
Therefore, the given system will have infinitely many solutions if k = 12.
kx − 2y + 6 = 0
4x + 3y + 9 = 0
kx − 2y + 6 = 0 4x + 3y + 9 = 0
Where, a_{1} = k, b_{1} = −2, c_{1} = 6
a_{2} = 4, b_{2} = −3, c_{2} = 9
Therefore, the given system of equations will have infinitely many solutions, if k = 8/3.
8x + 5y = 9
kx + 10y = 19
8x + 5y = 9 kx + 10y = 19
a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2 }=0
Where, a_{1} = 8, b_{1} = 5, c_{1 }= −9
a_{2} = k, b_{2} = 10, c_{2} = −18 a_{2} = k, b_{2} = 10,c_{2 }= −18
Therefore, the given system of equations will have infinitely many solutions, if k = 16.
2x − 3y = 7
(k + 2)x − (2k + 1)y = 3(2k − 1)
2x − 3y = 7 (k + 2)x − (2k + 1)y = 3(2k − 1)
Where, a_{1} = 2, b_{1} = −3, c_{1} = −7
a_{2 }= k, b_{2} = − (2k + 1), c_{2} = −3(2k − 1)
⇒ 2(2k + 1) = 3(k + 2) and 3 × 3(2k − 1) = 7(2k + 1)
⇒ 4k + 2 = 3k + 6 and 18k − 9 = 14k + 7
⇒ k = 4 and 4k = 16
⇒ k = 4
Therefore, the given system of equations will have infinitely many solutions, if k = 4.
2x + 3y = 2
(k + 2)x + (2k + 1)y = 2(k − 1)
2x + 3y = 2 (k + 2)x + (2k + 1)y = 2(k − 1)
Where, a_{1} = 2, b_{1} = 3,c_{1} = −2
a_{2} = (k + 2), b_{2} = (2k + 1),c_{2} = −2(k − 1)
⇒ 2(2k + 1) = 3(k + 2) and 3(k − 1) = (2k + 1)
⇒ 4k + 2 = 3k + 6 and 3k − 3 = 2k + 1
⇒ k = 4 and k = 4
x + (k + 1)y = 4
(k + 1)x + 9y = (5k + 2)
x + (k + 1)y = 4 (k + 1)x + 9y = (5k + 2)
Where, a_{1} = 1, b_{1} = (k + 1), c_{1} = −4
a_{2} = (k + 1), b_{2} = 9, c_{2} = − (5k + 2)
⇒ 9 = (k + 1)^{2} and (k + 1)(5k + 2) = 36
⇒ 9 = k^{2} + 2k + 1 and 5k^{2} + 2k + 5k + 2 = 36
⇒ k^{2 }+ 2k − 8 = 0 and 5k^{2} + 7k − 34 = 0
⇒ k^{2} + 4k − 2k − 8 = 0 and 5k^{2} + 17k − 10k − 34 = 0
⇒ k(k + 4) −2 (k + 4) = 0 and (5k + 17) − 2 (5k + 17) = 0
⇒ (k + 4)(k − 2) = 0 and (5k + 17)(k − 2) = 0
⇒ k = - 4 or k = 2 and k = -17/5 or k = 2
Thus, k = 2 satisfies both the condition.
Therefore, the given system of equations will have infinitely many solutions, if k = 2.
kx + 3y = 2k + 1
2(k + 1)x + 9y = (7k + 1)
kx + 3y = 2k + 1 2(k + 1)x + 9y = (7k + 1)
Where, a_{1 }= k, b_{1} = 3, c_{1} = −(2k + 1)
a_{2} = 2(k + 1), b_{2} = 9, c_{2} = −(7k + 1)
⇒ 9k = 3 × 2(k + 1) and 3(7k + 1) = 9(2k + 1)
⇒ 9k − 6k = 6 and 21k − 18k = 9 − 3
⇒ 3k = 6 ⇒ k = 2 and k = 2
2x + (k − 2)y = k
6x + (2k − 1)y = (2k + 5)
2x +( k − 2)y = k 6x + (2k − 1)y = (2k + 5)
Where, a_{1} = 2,b_{1} = (k − 2),c_{1} = −k
a_{2} = 6,b_{2} = (2k − 1),c_{2} = −(2k + 5)
⇒ 2k − 3k = −6 + 1 and k + k = 10
⇒ −k = −5 and 2k = 10 ⇒ k = 5 and k = 5
Therefore, the given system of equations will have infinitely many solutions, if k = 5.
2x + 3y = 72x + 3y = 7
(k + 1)x + (2k − 1)y = (4k + 1)
2x + 3y = 7 (k + 1)x + (2k − 1)y = (4k + 1)
a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2 }= 0
Where, a_{1} = 2, b_{1} = 3, c_{1} = −7
a_{2 }= k + 1, b_{2} = 2k − 1, c_{2} = −(4k + 1)
Extra close brace or missing open brace
⇒ 4k − 2 = 3k + 3 and 12k + 3 = 14k − 7
⇒ k = 5 and 2k = 10 ⇒ k = 5 and k = 5
2x + 3y = k
(k − 1)x + (k + 2)y = 3k
2x + 3y = k (k − 1)x + (k + 2)y = 3k
Where, a_{1 }= 2,b_{1 }= 3, c_{1} = −k
a_{2} = k − 1, b_{2 }= k + 2, c_{2} = −3k
⇒ 2k + 4 = 3k − 3 and 9 = k + 2 ⇒ 2k + 4 = 3k − 3 and 9 = k + 2 ⇒ k = 7 and k = 7
Therefore, the given system of equations will have infinitely many solutions, if k = 7.
Find the value of k for which the following system of equation has no solution:
kx − 5y = 2
6x + 2y = 7
kx − 5y = 2 6x + 2y = 7
Where, a_{1 }= k, b_{1} = −5, c_{1 }= −2
a_{2} = 6 b_{2 }= 2, c_{2} = −7
For no solution, we have
⇒ 2k = -30 ⇒ k = -15
Therefore, the given system of equations will have no solutions, if k = −15.
x + 2y = 0
2x + ky = 5
x_{2}y = 0 2x + ky = 5
Where, a_{1} = 1, b_{1} = 2, c_{1} = 0
a_{2} = 2, b_{2} = k, c_{2} = −5
Therefore, the given system of equations will have no solutions, if k = 4.
3x − 4y + 7 = 0
kx + 3y − 5 = 0
3x − 4y + 7 = 0 kx + 3y − 5 = 0
Where, a_{1} = 3, b_{1} = −4, c_{1} = 7
a_{2} = k, b_{2} = 3, c_{2} = −5
Therefore, the given system of equations will have no solutions, if k = - 9 /4.
2x − ky + 3 = 0
3x + 2y − 1 = 0
2x − ky + 3 = 0 3x + 2y − 1 = 0
Where, a_{1} = 2, b_{1} = −k, c_{1} = 3
a_{2} = 3, b_{2} = 2, c_{2} = −1
Therefore, the given system of equations will have no solutions, if k = – 4/3.
2x + ky − 11 = 0
5x − 7y − 5 = 0
2x + ky − 11 = 0 5x − 7y − 5 = 0
Where, a_{1} = 2, b_{1} = k, c_{1} = −11
a_{2} = 5, b_{2} = −7, c_{2} = −5a_{2} = 5, b_{2 }= −7, c_{2} = − 5
Therefore, the given system of equations will have no solutions, if k = -14/5.
kx + 3y = 3
12x + ky = 6
kx + 3y = 3 12x + ky = 6
Where, a_{1} = k, b_{1} = 3, c_{1} = −3
a_{2 }= 12, b_{2} = k, c_{2} = − 6
⇒ k^{2} = 36 ⇒ k = + 6 or −6
From (i)
Therefore, the given system of equations will have no solutions, if k = − 6.
For what value of a, the following system of equation will be inconsistent?
4x + 6y − 11 = 0
2x + ay − 7 = 0
4x + 6y − 11 = 0 2x + ay − 7 = 0
Where, a_{1} = 4, b_{1} = 6, c_{1} = −11
a_{2} = 2, b_{2} = a, c_{2} = −7
Therefore, the given system of equations will be inconsistent, if a = 3.
For what value of a, the following system of equation have no solution?
ax + 3y = a − 3
12x + ay = a
ax + 3y = a − 3 12x + ay = a
Where, a_{1} = a, b_{1 }= 3, c_{1} = - (a − 3)
a_{2} = 12, b_{2} = a, c_{2 }= − a
And,
⇒ a^{2 }= 36
⇒ a = + 6 or – 6?
a ≠ 6 ⇒ a = – 6
Therefore, the given system of equations will have no solution, if a = − 6.
Find the value of a, for which the following system of equation have
(i) Unique solution
(ii) No solution
kx + 2y = 5
3x + y = 1
kx + 2y − 5 = 0 3x + y − 1 = 0
Where, a_{1} = k, b_{1} = 2, c_{1} = −5
(i) For unique solution, we have
Therefore, the given system of equations will have unique solution, if k ≠ 6 k ≠ 6.
(ii) For no solution, we have
Therefore, the given system of equations will have no solution, if a = 6.
For what value of c, the following system of equation have infinitely many solution (where c ≠ 0 c ≠ 0)?
6x + 3y = c − 3
12x + cy = c
6x + 3y − (c − 3) = 0 12x + cy − c = 0
Where, a_{1} = 6, b_{1} = 3, c_{1} = −(c − 3)
a_{2 }= 12, b_{2} = c, c_{2} = - c
For infinitely many solution, we have
⇒ c = 6 and c – 3 = 3
⇒ c = 6 and c = 6
Therefore, the given system of equations will have infinitely many solution, if c = 6.
Find the value of k, for which the following system of equation have
(iii) Infinitely many solution
2x + ky = 1
3x − 5y = 7
2x + ky = 1 3x − 5y = 7
Where, a_{1} = 2, b_{1} = k, c_{1} = −1a_{1 }= 2, b_{1} = k, c_{1} = −1
a_{2 }= 3, b_{2} = −5, c_{2} = −7
Therefore, the given system of equations will have unique solution, if k ≠ -10/3.
Therefore, the given system of equations will have no solution, if k = -10)/3.
(iii) For the given system to have infinitely many solution, we have
So there is no value of k for which the given system of equation has infinitely many solution.
For what value of k, the following system of equation will represent the coincident lines?
x + 2y + 7 = 0
2x + ky + 14 = 0
x + 2y + 7 = 0 2x + ky + 14 = 0
Where, a_{1} = 1, b_{1} = 2, c_{1} = 7
a_{2} = 2, b_{2} = k, c_{2} = 14
The given system of equation will represent the coincident lines if they have infinitely many solution.
Therefore, the given system of equations will have infinitely many solution, if k = 4.
Find the value of k, for which the following system of equation have unique solution.
ax + by = c
lx + my = n
ax + by − c = 0 lx + my − n = 0
Where, a_{1} = a, b_{1} = b, c_{1 }= − c
a_{2 }= l, b_{2} = m, c_{2} = − n
Therefore, the given system of equations will have unique solution, if am ≠ bl.
Find the value of a and b such that the following system of linear equation have infinitely many solution:
(2a − 1)x + 3y − 5 = 0
3x + (b − 1)y − 2 = 0
The given system of equation may be written as,
(2a − 1)x + 3y − 5 = 0 3x + (b − 1)y − 2 = 0
Where, a_{1} = (2a − 1), b_{1} = 3, c_{1} = −5
a_{2} = 3, b_{2} = b − 1, c_{2} = −2
The given system of equation will have infinitely many solution, if
⇒ 2(2a − 1) = 15 and 6 = 5(b − 1)
⇒ 4a − 2 = 15 and 6 = 5b − 5 ⇒ 4a = 17 and 5b = 11
(a + b)x − (a + b − 3)y = 4a + b
2x − 3y − 7 = 0 (a + b)x − (a + b − 3)y − (4a + b) = 0
a_{2} = (a + b), b_{2} = −(a + b − 3), c_{2 }= −(4a + b)
⇒ 2(a + b − 3) = 3(a + b) and 3(4a + b) = 7(a + b − 3)
⇒ 2a + 2b − 6 = 3a + 3b and 12a + 3b = 7a + 7b − 21
⇒ a + b = −6 and 5a − 4b = −21
⇒ a = − 6 − b
Substituting the value of a in 5a − 4b = −21 we have
5( – b – 6) – 4b = – 21
⇒ − 5b − 30 − 4b = − 21
⇒ 9b = − 9 ⇒ b = −1
As a = – 6 – b
⇒ a = − 6 + 1 = − 5
Hence the given system of equation will have infinitely many solution if
a = – 5 and b = –1.
Find the value of p and q such that the following system of linear equation have infinitely many solution:
2x − 3y = 9
(p + q)x + (2p − q)y = 3(p + q + 1)
2x − 3y − 9 = 0 (p + q)x + (2p − q)y − 3(p + q + 1) = 0
Where, a_{1} =2, b_{1} = 3, c_{1 }= −9
a_{2} = (p + q), b_{2} = (2p − q), c_{2} = -3(p + q + 1)
2(2p - q) = 3(p + q) and (p + q + 1) = 2p - q
⇒ 4p - 2q = 3p + 3q and -p + 2q = -1
⇒ p = 5q and p - 2q = 1
Substituting the value of p in p - 2q = 1, we have
3q = 1
⇒ q = 1/3
Substituting the value of p in p = 5q we have
p = 5/3
p = 5/3 and q = 1/3.
Find the values of a and b for which the following system of equation has infinitely many solution:
(i) (2a − 1)x + 3y = 5
3x + (b − 2)y = 3
(ii) 2x − (2a + 5)y = 5
(2b + 1)x − 9y = 15
(iii) (a − 1)x + 3y = 2
6x + (1 − 2b)y = 6
(iv) 3x + 4y = 12
(a + b)x + 2(a − b)y = 5a – 1
(v) 2x + 3y = 7
(a − 1)x + (a + 1)y = 3a − 1
(vi) 2x + 3y = 7
(a − 1)x + (a + 2)y = 3a
(i) The given system of equation may be written as,
(2a − 1)x + 3y − 5 = 0 3x + (b − 2)y − 3 = 0
Where, a_{1} = 2a − 1, b_{1 }= 3, c_{1} = −5
a_{2} = 3, b_{2} = b − 2, c_{2 }= -3(p + q + 1)
2a - 1 = 5 and - 9 = 5(b - 2)
⇒ a = 3 and -9 = 5b - 10
a = 3 and b = 1/5
a = 3 and b = 1/5.
(ii) The given system of equation may be written as,
2x − (2a + 5)y = 5 (2b + 1)x − 9y = 15
Where, a_{1} = 2, b_{1} = - (2a + 5), c_{1} = −5
a_{2} = (2b + 1), b_{2} = −9, c_{2} = −15
a = - 1 and b = 5/2.
(iii) The given system of equation may be written as,
(a − 1)x + 3y = 2 6x + (1 − 2b)y = 6
Where, a_{1} = a-1, b_{1} = 3, c_{1} = −2
a_{2} = 6, b_{2} = 1 − 2b, c_{2} = −6
⇒ a – 1 = 2 and 1 – 2b = 9
⇒ a - 1 = 2 and 1 - 2b = 9
⇒ a = 3 and b = -4
a = 3 and b = −4.
(iv) The given system of equation may be written as,
3x + 4y − 12 = 0 (a + b)x + 2(a − b)y − (5a − 1) = 0
Where, a_{1} = 3, b_{1} = 4, c_{1} = −12
a_{2 }= (a + b), b_{2} = 2(a − b), c_{2} = – (5a − 1)
⇒ 3(a - b) = 2a + 2b and 2(5a - 1) = 12(a - b)
⇒ a = 5b and -2a = -12b + 2
Substituting a = 5b in -2a = -12b + 2, we have
-2(5b) = -12b + 2
⇒ −10b = −12b + 2 ⇒ b = 1
Thus a = 5
a = 5 and b = 1.
(v) The given system of equation may be written as,
2x + 3y − 7 = 0 (a − 1)x + (a + 1)y − (3a − 1) = 0
a_{2} = (a − 1), b_{2} = (a + 1), c_{2} = - (3a − 1)
⇒ 2(a + 1) = 3(a - 1) and 3(3a - 1) = 7(a + 1)
⇒ 2a - 3a = -3 - 2 and 9a - 3 = 7a + 7
⇒ a = 5 and a = 5
(vi) The given system of equation may be written as,
2x + 3y − 7 = 0 (a − 1)x + (a + 2)y − 3a = 0
Where, a_{1 }= 2, b_{1} = 3, c_{1} = −7
a_{2} = (a − 1), b_{2 }= (a + 2), c_{2} = −3a
⇒ 2(a + 2) = 3(a - 1) and 3(3a) = 7(a + 2)
⇒ 2a + 4 = 3a - 3 and 9a = 7a + 14
⇒ a = 7 and a = 7
a = 7and b = 1.
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Chapter 3: Pair of Linear Equations in Two...