Find the values of a and b for which the following system of linear equations has infinitely many solutions:
kx + 3y – (k – 3) = 0, 12x + ky – k = 0
Given, pair of equations
kx + 3y – (k – 3) = 0
and 12x + ky – k = 0
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = k, b1 = 3 and c1 = – (k – 3)
and a2 = 12, b2 = k and c2 = – k
For infinitely many solutions,
Here, 
∴ 
…(1)
On taking I and II terms, we get
⇒ k2 = 36
⇒ k = √36
⇒ k = ±6
But k = – 6 not satisfies the last two terms of eqn (1)
On taking II and III terms, we get
⇒ 3k = k(k – 3)
⇒ 3k = k2 – 3k
⇒ k2 – 3k – 3k = 0
⇒ k(k – 6) = 0
⇒ k = 0 and 6
Which satisfies the last two terms of eqn (1)
Hence, the required value of k = 0, 6
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