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,
$(k + 2) x + 10 y = k$
$k x + (k + 3) y = k - 1$ has no solution is :

A
infinitely many
B
1
C
2
D
3

Solution:

Given equations will have no solution if

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

On solving first two, we have

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

For $k = 2$ these equations are identical.

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

$∴$ There is only one value of k.

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