CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

For what value of k does the system of equations x+2y=5,3x+ky+15=0 have (i) a unique solution,(ii) no solution?


Solution

The equations are written as
x+2y-5=0
3x+ky+15=0

a1 = 1  b1 = 2  c1 = 5
a2 = 3  b2 = k  c2 = 15

(i) For unique solution,we have

a1a2 b1b2
132k

k≠2×3⇒k≠6

Therefore, the given system will have a unique solution for all real values of k other than 6.

(ii) For no solution, we have

a1a2 =b1b2c1c2
13515
13=2k
k=6

Therefore, the given system of equations will have no solutions, if k=6

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More...


Same exercise questions
QuestionImage
QuestionImage
View More...



footer-image