CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Find the value of k for which the given system of equations has infinite number of solutions.
5x+2y=2k and 2(k+1)x+ky=(3k+4)

A
4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
7
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 4
Comparing 5x+2y=2k with a1x+b1y+c1=0, we get,
a1=5,b1=2,c1=2k

Comparing 2(k+1)x+ky=3k+4 with a2x+b2y+c2=0, we get,
a2=2(k+1),b2=k,c2=(3k+4)

Now,
Given: the equations has infinite no of solutions.
a1a2=b1b2=c1c2

52(k+1)=2k=2k(3k+4)

When,

52(k+1)=2k

5k=4(k+1)
5k4k=4
k=4

and

2k=2k(3k+4)

2(3k+4)=2k2
3k+4=k2
k23k4=0
k24k+k4=0
k(k4)+(k4)=0
(k4)(k+1)=0

Either, k=4 or k=1
k=4 is the value that satisfies both the conditions.

So, Option A is correct

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Graphical Solution
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon