# The system x – 2y = 3 and 3x + ky = 1 has a unique solution only whenA. k = – 6B. k ≠ – 6C. k = 0D. k ≠ 0

We have,

x – 2y – 3 = 0

3x + ky – 1 = 0

The given equation is in the form: a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

Here, we have:

a1 = 1, b1 = – 2 and c1 = – 3

And, a2 = 3, b2 = k and c2 = – 1 , and These graph lines will intersect at a unique point when we have:   Hence, k has all real values other than – 6

Thus, option B is correct

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Dealing With the Real Life Problems53 mins  Quiz | Real Life Problems Through Linear Equations56 mins  Quiz | Solution of Linear Equations53 mins  Champ Quiz | Consistency and Inconsistency of Solutions36 mins  Pair of Linear Equations in Two Variables46 mins  Elimination (quicker than quickest)44 mins  Consistent and Inconsistent Equations33 mins  Master Substitution Method46 mins  Real Life Problems Through Linear Equations41 mins  All Kinds of Word Problems in Linear Equations42 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 