For what value of k, 2x + 3y = 4 and (k + 2)x + 6y = 3k + 2 will have infinitely many solutions.
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Answered by
113
a1/a2=b1/b2=c1/c2
2/(k+2)=3/6=4/3k+2
we take
2/(k+2)=3/6
(cross multiplication)
3(k+2)=6(2)
3k +6=12
3k =12-6
3k =6
k=3/6
k=2
2/(k+2)=3/6=4/3k+2
we take
2/(k+2)=3/6
(cross multiplication)
3(k+2)=6(2)
3k +6=12
3k =12-6
3k =6
k=3/6
k=2
Answered by
50
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