find the value of k for which the system of equatio has infinitely many solution 2x+(k-2)y=k and 6x+(2k-1)y=2k+5
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we know that condition for infinitely many solutions is:-
a1/a2= b1/b2= c1/c2
by applying this condition we get:-
2/6= (k-2)/(2k-1)= k/(2k+5)
to find k we take:-
2/6= k/(2k+5), by cross-multiplying we get:-
4k+10= 6k
4k+10-6k= 0
-2k+10= 0
-2k= -10
2k= 10 (because minus gets cancelled in rhs and lhs)
k= 10/2
k= 5
Therefore the value of k is 5.
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a1/a2= b1/b2= c1/c2
by applying this condition we get:-
2/6= (k-2)/(2k-1)= k/(2k+5)
to find k we take:-
2/6= k/(2k+5), by cross-multiplying we get:-
4k+10= 6k
4k+10-6k= 0
-2k+10= 0
-2k= -10
2k= 10 (because minus gets cancelled in rhs and lhs)
k= 10/2
k= 5
Therefore the value of k is 5.
Hope this answer helps you!
if it helps you so plz mark it brainliest
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