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(D) The given equations are:$x - ky - 2 = 0$ ...... $(1)$$3x + 2y + 5 = 0$ ...... $(2)$Comparing these with the general form $a_1x + b_1y + c_1 = 0$ a

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$-\frac{2}{3}$

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(D) The given equations are:$x - ky - 2 = 0$ ...... $(1)$$3x + 2y + 5 = 0$ ...... $(2)$Comparing these with the general form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$a_1 = 1, b_1 = -k, c_1 = -2$$a_2 = 3, b_2 = 2, c_2 = 5$For a unique solution,the condition is $\frac{a_1}{a_2} 
eq \frac{b_1}{b_2}$.Substituting the values:$\frac{1}{3} 
eq \frac{-k}{2}$Multiplying both sides by $6$:$2 
eq -3k$$k 
eq -\frac{2}{3}$Thus,the system has a unique solution for all real values of $k$ except $k = -\frac{2}{3}$.

If the following pair of equations is consistent (unique solution),then find the value of $k$:
$x - ky = 2$
$3x + 2y = -5$

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If the following pair of equations is consistent (unique solution),then find the value of $k$:$x - ky = 2$$3x + 2y = -5$