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(A) The given equations are:$3x - 4y + 7 = 0$  ... $(1)$$kx + 3y - 5 = 0$  ... $(2)$Comparing these with the general form $a_1x + b_1y + c_1 = 0$ and

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$-\frac{9}{4}$

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(A) The given equations are:$3x - 4y + 7 = 0$  ... $(1)$$kx + 3y - 5 = 0$  ... $(2)$Comparing these with the general form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$a_1 = 3, b_1 = -4, c_1 = 7$$a_2 = k, b_2 = 3, c_2 = -5$For a pair of linear equations to have no solution,the condition is:$\frac{a_1}{a_2} = \frac{b_1}{b_2} 
eq \frac{c_1}{c_2}$Substituting the values:$\frac{3}{k} = \frac{-4}{3} 
eq \frac{7}{-5}$Since $\frac{-4}{3} 
eq \frac{7}{-5}$,we only need to solve for:$\frac{3}{k} = \frac{-4}{3}$Cross-multiplying gives:$-4k = 9$$k = -\frac{9}{4}$Thus,the value of $k$ is $-\frac{9}{4}$.

If the following pair of equations does not have any solution,then find the value of $k$:
$3x - 4y + 7 = 0$
$kx + 3y - 5 = 0$

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If the following pair of equations does not have any solution,then find the value of $k$:$3x - 4y + 7 = 0$$kx + 3y - 5 = 0$