Find the value of `k` for which the following system of linear equations has infinite solutions: `x+(k+1)y=5,\ \ \ \ (k+1)x+9y=8k-1`

Explanation

The given system of equations is
` x + ( k + 1 ) y - 5 = 0 " " `… (i)
`(k + 1) x + 9y + (1 - 8k ) = 0 " " `… (ii)
These equations are of the form
` a_ 1 x + b _ 1 y + c_ 1 = 0 and a _ 2 x + b_ 2 y + c _ 2 = 0 `
where ` a_ 1 = 1 , b_ 1 = ( k + 1 ), c_ 1 = - 5`
and ` a_ 2 = ( k + 1 ), b_ 2 = 9, c_ 2 = ( 1 - 8k )`.
` therefore ( a_ 1 ) /( a_ 2) = (1)/(( k +1)) , (b _ 1 ) /( b_ 2 ) = (( k + 1 ) )/(9 ) and (c _ 1 ) /(c _ 2) = (-5)/(( 1- 8k )) = ( 5)/(( 8k - 1 ))`
Let the given system of equations have infinitely many solutions.
Then, ` (a_ 1 )/(a_ 2 ) = (b_ 1)/(b_ 2 ) = (c_ 1 )/(c_ 2)`
` rArr ( 1)/(( k + 1 )) = ((k + 1 ))/( 9) = ( 5)/(( 8 k - 1 )) `
` rArr (1)/(( k + 1 )) = ((k + 1 ) )/(9) and (( k +1 ))/( 9 ) = ( 5)/( ( 8 k - 1))`
` rArr ( k + 1 ) ^(2) = 9 and ( k + 1 ) ( 8 k - 1 ) = 45`
` rArr ( k + 1 = 3 or k + 1 = - 3 ) and 8 k ^(2) + 7k - 46 = 0 `
` rArr ( k = 2 or k = - 4 ) and ( k - 2 ) ( 8k + 23)= 0 `
` rArr ( k = 2 or k = - 4) and ( k = 2 or k = (-23)/(8)) `
`rArr k = 2`.
Hence, the given system of equations will have infinitely many solutions when k = 2.