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

Answer 1

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Answer 2

Solution:

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Given :

the system of equations: (k+1)x+8y=4k &  kx+(k+3)y=3k-1 has no solution,.

(k+1)x + 8y - 4k = 0 &  kx + (k+3)y - 3k + 1 = 0
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To find :

The value of k,.
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We know that,

 The equations are in the form,

$a_1x + b_1y + c_1 = 0$

&

$a_2x + b_2y + c_2 = 0$

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As, the equations are inconsistent,.

We can say that,.

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$


Hence,

⇒ $\frac{k + 1}{k} = \frac{8}{k+3}$

⇒ $(k + 1)(k + 3) = (8)(k)$

⇒ k² + 4k + 3 = 8k

⇒ k² + 4k + 3 - 8k = 0

⇒ k² - 4k + 3 = 0

⇒ (k - 3)(k - 1) = 0

For the equation to be 0,

Either,

k - 3 = 0  (or) k - 1 = 0

k = 3 (or) k = 1,.

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