The pair of linear equations (3k+1)x+3y-5=0 and 2x-3y+5=0 have infinite solutions.Then find the value of k.
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Hi!
The answer is:-
3k+1/2=3/-3=-5/5
3k+1/2=3/-3
-9k-3=6
-9k=6+3
-9k=9
k=-1
and
3k+1/2=-5/5
15k+5=-10 (By cross multipication)
15k=-15
k=-1
The answer is:-
3k+1/2=3/-3=-5/5
3k+1/2=3/-3
-9k-3=6
-9k=6+3
-9k=9
k=-1
and
3k+1/2=-5/5
15k+5=-10 (By cross multipication)
15k=-15
k=-1
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