Linear Inequalities

Question
CBSEENMA11015487

The number of values of k, for which the system of equations
(k+1) x + 8y = 4k
kx + (k+3)y = 3k -1
has no solution, is 

  • infinite 

  • 1

  • 2

  • 3

Solution

B.

1

Condition for the system of equations has no solution,
straight a subscript 1 over straight a subscript 2 space equals straight b subscript 1 over straight b subscript 2 space not equal to straight c subscript 1 over straight c subscript 2
therefore space fraction numerator straight k plus 1 over denominator straight k end fraction space equals space fraction numerator 8 over denominator straight k plus 3 end fraction space not equal to fraction numerator 4 straight k over denominator 3 straight k minus 1 end fraction
Take space fraction numerator straight k plus 1 over denominator straight k end fraction space equals space fraction numerator 8 over denominator straight k plus 3 end fraction
rightwards double arrow space straight k squared space plus space 4 straight k space plus 3 space equals space 8 straight k
rightwards double arrow straight k squared minus 4 straight k space plus 3 space equals space 0
rightwards double arrow left parenthesis straight k minus 1 right parenthesis left parenthesis straight k minus 3 right parenthesis space space equals 0
straight k space equals space 1 comma 3
If space straight k space equals 1 comma space then space fraction numerator 8 over denominator 1 plus 3 end fraction space equals space fraction numerator 4.1 over denominator 2 end fraction comma space false
Therefore, k = 3
Hence, only one value of k exists.

Sponsor Area

Some More Questions From Linear Inequalities Chapter