Math, asked by sahilkund42, 10 months ago

Find k so that the following pair of linear equations has no solution.
3x + y = 1; (2k – 1)x + (k - 1)y = 2k + 1​

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Answers

Answered by Anonymous
3

Hope this helps you. .

- Khushali.S.S. .

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Answered by Priyanshulohani
0

\underline\mathfrak{Given:-}

\: \: \: \: \: \: \: {({2k} \: - \: {1})} \: x \: + \: {({k} \: - \: {2})} \: y \: \: = \: \: {5}

\: \: \: \: \: \: \: {({k} \: + \: {2})} \: x \: + \: y \: \: = \: \: {3}

\underline\mathfrak{To \: \: Find:-}

\: \: \: \: \: The \: \: value \: \: k \: ?

\underline\mathfrak{Solutions:-}

\: \: \: \: \: \fbox{\dfrac{a_1}{a_2} \: \: = \: \:  \dfrac{b_1}{b_2} \: \: \neq \: \: \dfrac{c_1}{c_2}}

\: \: \: \: \: \dfrac{{2k} \: - {1}}{{k} \: + \: {2}} \: \: = \: \:  \dfrac{{k} \: - \: {2}}{{1}} \: \: \neq \: \: \dfrac{5}{3}

\: \: \: \: \: \leadsto \dfrac{{2k} \: - {1}}{{k} \: + \: {2}} \: \: = \: \:  \dfrac{{k} \: - \: {2}}{{1}} \: \: \: \: \: .....{(1)}.

\: \: \: \: \: \leadsto \dfrac{{k} \: - \: {2}}{{1}} \: \: \neq \: \: \dfrac{5}{3} \: \: \: \: \: .....{(2)}.

\: \: \: \: \: Now, \: \: cross \: \: multiple \: \: in \: \: Eq. \: \: {(1)}.

\: \: \: \: \: \leadsto \dfrac{{2k} \: - {1}}{{k} \: + \: {2}} \: \: = \: \:  \dfrac{{k} \: - \: {2}}{{1}}

\: \: \: \: \: \leadsto {{2k} \: - {1}} \: \: = \: \: {({k} \: - \: {2})} \: \times \: {{({k} \: + \: {2})}}

\: \: \: \: \: \leadsto {{2k} \: - {1}} \: \: = \: \: {{k}^{2} \: - \: {2}^{2}} \: \: \: \: \: \: \: \: \: {[(a \: + \: b) \: (a \: - \: b) \: \: = \: \: ({a}^{2} \: - \: {b}^{2}]}

\: \: \: \: \: \leadsto {{2k} \: - {1}} \: \: = \: \: {{k}^{2} \: - \: {4}}

\: \: \: \: \: \leadsto {0} \: \: = \: \: {k}^{2} \: - \: {2k} \: - \: {4} \: + \: {1}

\: \: \: \: \: \leadsto {0} \: \: = \: \: {k}^{2} \: - \: {2k} \: - \: {3}

\: \: \: \: \: \leadsto {k}^{2} \: - \: {2k} \: - \: {3}

\: \: \: \: \: \leadsto {k} \: {({k} \: - \: {3})} \: + \: {1} \: {({k} \: - \: {3})}

\: \: \: \: \: \leadsto {({k} \: + \: {1})} \: \: \: {({k} \: - \: {3})}

\: \: \: \: \: \leadsto {k} \: \: = \: \: {-1} \: \: \: Or \: \: \: {k} \: \: = \: \: {3}

\: \: \: \: \: Hence, \: \: the \: \: the \: \: value \: \: of \: \: k \: \: is \: \:{-1} \: \: and \: \: {3}.

\: \: \: \: \:  \dfrac{{k} \: - \: {2}}{{1}} \: \: \neq \: \: \dfrac{5}{3} \: \: \: \: \: .....{(2)}.

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