15. Find the value of k for which pair of linear equations 3x + 2y = -

1.	15. Find the value of k for which pair of linear equations 3x + 2y = -5 and x - ky =2 has a unique solution
Answer» Answer: $\bigstar{\bold{k\neq \dfrac{-2}{3} , k \in R}}$ Step-by-step explanation: $\Large{\underline{\sf{Given:}}}$ A pair of linear EQUATIONS: 3x + 2y = -5 x - ky = 2 $\Large{\underline{\sf{To\:Find:}}}$ The value of k so that the equations have a unique SOLUTION $\Large{\underline{\sf{Solution:}}}$ ➟ Here we have to find the value of k so that the given pair of equations is consistent and has a unique solution. ➟ If a pair of equations are consistent, we know that, $\tt \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$ ➟ Here a₁ = 3 a₂ = 1 b₁ = 2 b₂ = -k ➟ Substituting the datas we get, $\tt \dfrac{3}{1} \neq \dfrac{2}{-k}$ ➟ Cross multiplying we get, -3K ≠ 2 k ≠ -2/3 ➟ Hence k can take the value of any real number EXCEPT -2/3. $\boxed{\bold{k\neq \dfrac{-2}{3} , k \in R}}$ $\Large{\underline{\sf{Notes:}}}$ ➟ If a pair of linear equations is: ➵ consistent and has a unique solution then, $\tt \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$ ➵ consistent and has infinite number of solutions, $\tt \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$ ➵ inconsistent and has no solution, $\tt \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}\neq \dfrac{c_1}{c_2}$