Question

For what value of K will the following
pair of linear equation have no
solution ? x+2y=3. (k-1)x+(k+1)y=k+2​

Answer 1

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The equation are

x + 2y − 3 = 0     ....(1)

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

Here, a1=1 , b1=2 , c1=−3

and a2=k-1 , b2=k+1 ,c2= - ( k+2 )

The   given   system   has   no   solution   if

$\frac{a1}{a2} = \frac{b1}{b2} \: not \: equal \: \frac{c1}{c2}$

$\frac{1}{k - 1} = \frac{2}{k + 1}$

k+1 = 2( k-1 )

k+1 = 2k - 2

1+2 = 2k -k

3 = k

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