The number of values of k, for which the system of equations: (k + 1) x + 8y = 4k kx + (k + 3) y = 3k

Engineering

Mathematics

Solving System of Linear Equation Cramers Rule

Question

The number of values of k, for which the system of equations:

(k + 1) x + 8y = 4k

kx + (k + 3) y = 3k – 1

has no solution, is :

Infinite

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For no solution

$\frac{(k + 1)}{k} = \frac{8}{k + 3} \neq \frac{4 k}{3 k - 1}$

$\Rightarrow$ From 1^st and 2^nd ratio

(k + 1) (k + 3) = 8k

$\Rightarrow$ k² + 4k + 3 = 8k

$\Rightarrow$ k = 1, 3

For k = 1,

$\frac{8}{k + 3} = 2, \frac{4 k}{3 k - 1} = 2$

$\Rightarrow$ k = 1 not possible

For k = 3,

$\frac{8}{k + 3} = \frac{4}{3}, \frac{4 k}{3 k - 1} = \frac{3}{2}$

$\Rightarrow$ k = 3 is possible.