for which value of k following pair of linear equationhave no solution 3k+y=1 (2k-1) x +(k-1) y=2k+1​

Question

for which value of k following pair of linear equationhave no solution 3k+y=1 (2k-1) x +(k-1) y=2k+1​

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Maya 2 weeks 2021-10-01T20:12:30+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-01T20:13:51+00:00

    \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)}.

    __________________________________

    0
    2021-10-01T20:14:28+00:00

    Answer:

    3x + y =1(2k-1)x+(k-1)y=2k+1

    => 3x+y-1=0

    (2k-1)x+(k-1)y-2k-1 =0

    We know,

    a1/a2 = b1/b2 not equal to c1/c2 has no solution

    Now we have,

    a1=3 , b1=1

    a2=2k-1 , b2 = k-1

    3/2k-1 = 1/k-1

    3(k-1) = 2k -1

    3k-3 = 2k -1

    3k -2k = -1 +3

    k = 2

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