Math, asked by rushilgaba98, 1 year ago

find the value of k for which the following pair of linear equations has infinite solutions. 3x +y=2 , (2k-1)X+(k-1)y=2k-1​

Answers

Answered by rohanjaat73
1

Step-by-step explanation:

kx+3y_(k_3) =0

12x+ky_k =0

a1/a2=k/12,b1/b2 =3/k

for a pair of linear equations to have inf Infinitely many solutions . a1/a2 =b1/b2=c1/c2 .

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)}.

__________________________________

Similar questions