# 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.

There are two equations given in the question:

2x + 3y – 7 = 0 (i)

And, (k – 1)x + (k + 2)y – 3k = 0 (ii)

These given equations are in the form a1x + b1y + c1 = 0 and

a2x + b2y + c2 = 0 where,

a1 = 2, b1 = 3 and c1 = – 7

Also, a2 = (k – 1), b2 = (k + 2) and c2 = – 3k

Now, for the given pair of linear equations having infinitely many solutions we must have:

, and

2 (k + 2) = 3 (k – 1), 3 × 3k = 7 (k + 2) and 2 × 3k = 7 (k – 1)

2k + 4 = 3, 9k = 7k + 14 and 6k = 7k – 7

k = 7, k = 7 and k = 7

Hence, the value of k is 7

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