# 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