The number of values of k for which the system of linear equations,(k+2)x+10y=kkx+(k+3)y=k−1 has no solution is :

The number of values of k for which the system of linear equations,$\left(k+2\right)x+10y=k$$kx+\left(k+3\right)y=k-1$ has no solution is :

1. A

infinitely many

2. B

1

3. C

2

4. D

3

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Solution:

Given equations will have no solution if

$\frac{k+2}{k}=\frac{10}{k+3}\ne \frac{k}{k-1}$

On solving first two, we have

$\left(k+2\right)\left(k+3\right)=10k⇒{k}^{2}-5k+6=0⇒k=2,3$

For $k=2$ these equations are identical.

So, for only k = 3 these equations have no solution

$\therefore$There is only one value of k.

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