Question

Need answer.....
Q 20

Answer 1

the condition for infinite number of solutions is

a1/a2 = b1/b2 = c1/c2

a1 = 2      a2 = (k+2)

b1 = 3      b2 = (2k +1)

c1 = (k+2)   c2 = 2 ( k-1)

so

2/(k+2) = 3/ (2k +1) = (k+2)/2 ( k-1)

taking the first two

2/(k+2) = 3/(2k+1)

cross multiply

4 k + 2 = 3k + 6

k = 4

therefore the value of k is 4

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