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The value of $k$ for which the system $kx + 3 y = 7$ and $2 x - 5 y = 3$ has no solution is

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7

and

k = - \frac{3}{14}

4

and

k = \frac{3}{14}

\frac{6}{5}

and

k \neq \frac{14}{3}

- \frac{6}{5}

and

k \neq \frac{14}{3}

answer is D.

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Detailed Solution

It is given the system of linear equation,

kx + 3 y = 7

and

2 x - 5 y = 3

.
Now comparing the given equations with the system of linear equations ,

a_{1} x + b_{1} y

c_{1}

= 0…(1)

a_{2} x

b_{1} y + c_{1} = 0

….(2) we get,

a_{1} = k_{1}, b_{1} = 3, c_{1} = - 7

a_{2} = 2, b_{2} = - 5, c_{2} = - 3

For no solution the system of linear equations must satisfy,

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

\Rightarrow

\frac{k}{2} = \frac{3}{- 5} \neq \frac{- 7}{- 3}

Now solving the above first two equality we get,

\Rightarrow \frac{k}{2} = \frac{3}{- 5}

\Rightarrow

k = \frac{- 6}{5}

Now taking the last two equality we get,

\Rightarrow \frac{k}{2} \neq \frac{- 7}{- 3}

\Rightarrow

k \neq \frac{14}{3}

So the correct option is 4.

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