Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts Dedicated counsellor for each student Detailed Performance Evaluation view all coursesPage 2Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts Dedicated counsellor for each student Detailed Performance Evaluation view all coursesThe given system of equation is 2x + ky - 11 =0 5x − 7y - 5 = 0 The system of equation is of the form `a_1x + b_1y + c_1 = 0` `a_2x + b_2y + c_2 = 0` Where, `a_1 = 2,b_1 = k, c_1 = -11` And `a_2 = 5, b_2 = -7, c_2 = -5` For a unique solution, we must have `a_1/a_2 - b_1/b_2 != c_1/c_2` `=> 2/5 = k/(-7) != (-11)/(-5)` Now `2/5 = k/(-7)` `=> 2 xx (-7) = 5k` `=> 5k = -14` `=> k = (-14)/5` Cleary for `(-14)/5` we have `k/(-7) != (-11)/(-5)` Hence, the given system of equation will have no solution if `k = (-14)/5` Page 2Find the value of k for which the following system of equations has a unique solution: kx + 3y = 3 12x + ky = 6 kx + 3y = 3 12x + ky = 6 For no solution `a_1/a_2 - b_1/b_2 != c_1/c_2` `=> k/12 = 2/k != 3/6` `k/12 = 3/k` `k^2 = 36` `k = +- 6` i.e k = 6, -6 Also `3/k != 3/6` `(3 xx 6)/3 != k` `k != 6` k = -6 satisfies both the condition Hence k = -6 Concept: Pair of Linear Equations in Two Variables Is there an error in this question or solution? Page 3The given system of equation may be written as 4x + 6y - 11 = 0 2x + ky - 7 = 0 The system of equation is of the form `a_1x + b_1y + c_1 = 0` `a_2x + b_2y + c_2 = 0` Where `a_1 = 4, b_1 = 6, c_1 = -11` And `a_2 = 2, b_2 = k, c_2 = -7` For a unique solution, we must have `a_1/a_2 = b_1/b_2 != c_1/c_2` Now `a_1/a_2 = b_1/b_2` `=> 4/2 = 6/k` `=> 4k = 12` `=> k = 12/4 = 3` Clearly, for this value of k, we have `a_1/a_2 = b_1/b_2 != c_1/c_2 Hence, the given system of equation is inconsistent, when k = 3 Page 4The given system of the equation may be written as αx + 3y -α - 3 = 0 12x + αy - α = 0 The system of equation is of the form `a_1x + b_1y + c_1 = 0` `a_2x + b_2y + c_2 = 0` Where `a_1 = α, b_1 = 3, c_1 = -(α - 3)` And `a_2 = 12, b_2 = α, c_2 = -α` For a unique solution, we must have
`=> α/12 = 3/α != (-(α - 3))/(-α)` Now, `3/α != (-(α - 3))/(-α)` `=> 3/α != (α - 3)/α` `=> 3 != α - 3` `=> 3 = 3 != α` `=> 6 != α` `=> α != 6` And `α/12 = 3/α` `=> α^2 = 36` `=> α +- 6` `=> α = -6` [∵ `α != 6`] Hence, the given system of equation will have no solution if α = -6 |