Question

find the value of k for which the pair of linear equation 2 X + 3 Y = 27 and K - 1 X + K + 2 Y is equal to 3 K has infinitely many solutions of the problem ​

Answer 1

Answer:

As a1/a2 is not equal to b1/b2

Therefore there is no value of k for which the given pair of equation has infinitely many solution

Answer 2

$\bold{Given:-} \begin{cases} \sf{a_1 = 2.} \\ \sf{a _ 2 = k - 1} \\ \sf{b _ 1 = 3.} \\ \sf{b_2 = k + 2.} \\ \sf{c</p><p>_1 = 27.} \\ \sf{c _2 = 3k.} \end {cases}$

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$\underline \bold{Solution:-}$

$Have \: \infty \: Solution.$

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

$2x + 3y = 27. - - - - - - - - (1) \\ \\ (k + 1)x +( k + 2)y = 3k - - - ( 2)$

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$Putting \: value:-$

$\frac{2}{(k - 1)} = \frac{3}{(k + 2)} = \frac{{ \cancel-} 27}{ { \cancel -} 3k} .$

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$Taking :-$

$\frac{2}{(k - 1)} = \frac{3}{(k + 2)}$

$\implies \: 2(k + 2) = 3(k - 2). \\ \\ \implies \: 2k + 4 = 3k - 3. \\ \\ \implies \: k = 7.$

But,

It not have infinity many solution.

because the value of k not be equal all the case.