Question

For what valu of k the following pair of linear equations have no solution:
3x+y=1
(2k-1)x - (k-1)y= 2k+1

Answer 1

Polynomials,

We have the pair of linear equations as -

3x + y = 1.......(1)
and,(2k - 1)x - (k -1)y = 2k + 1.......(2)

Here,
Comparing by a1x + b1y = c1
and a2x + b2y = c2 we get

a1 = 3, b1 = 1 and c1 = 1
also, a2 = (2k -1), b2 = -(k - 1) and c2 = (2k +1)

So the pair of linear equations will have no solution when,
a1/a2 = b1/b2 not= c1/c2

Now,
a1/a2 = 3/(2k - 1) b1/b2 = 1/{-(k - 1)}
So to have no solution it must be a1/a2 = b1/b2

Therefore,
3/(2k - 1) = 1/{-(k - 1)}
or, 3/(2k - 1) = -1/(k - 1)
or, 3k -3 = -2k +1
or, 5k = 4
or, k = 4/5

Now placing the values of k we get a1/a2 = b1/b2 but not = c1/c2

So the pair of linear equations will have no solution when k = 4/5.

That's it
Hope it helped.

Answer 2

$\underline\mathfrak{Given:-}$

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

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

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

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

$\underline\mathfrak{Solutions:-}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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