If am=bl, then find whether the pair of linear equations ax+by=c and lx+my=n has no solution,unique solution or infinitely many solutions.
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Hey
Given :-
am = bl
=> a / l = b / m
Now , equations :-
ax + by - c = 0
lx + my - n = 0
Now ,
a / l = b / m =/= -c / -n
so a1 / a2 = b1 / b2 =/= c1 / c2
It is no solution .
thanks :)
Given :-
am = bl
=> a / l = b / m
Now , equations :-
ax + by - c = 0
lx + my - n = 0
Now ,
a / l = b / m =/= -c / -n
so a1 / a2 = b1 / b2 =/= c1 / c2
It is no solution .
thanks :)
preetpragyan:
Thank you
mark brainliest if it helped :)
But how can u say b/m is not equal to -c/-n
bcz it is given that a / l = b / m ....but not given about c/n and it's a variable !! So
But that doesn't mean c/n is not equal to it
yes that's the point to be understood !! If given was not there then it would be the case of unique solution ...!! That means what i have mentioned
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