Question

Find the value of k for which the following pair of linear equations have infinitely many solutions:

2x + 3y = 7, (k – 1)x + (k + 2)y = 3k

Answer 1

2x + 3y – 7 = 0 and (k – 1)x + (k + 2)y – 3k = 0.

For infinitely many solutions, we have `(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)`.

∴ `2/((k - 1)) = 3/((k + 2))` and `3/((k + 2)) = 7/(3k)`

⇒ 2k + 4 = 3k – 3 and 9k = 7k + 14 

⇒ k = 7