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 infinite 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