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Last updated at Aug. 2, 2021 by Teachoo

Example 16 For what values of k will the following pair of linear equations have infinitely many solutions? kx + 3y – (k-3) = 0 12x + ky – k = 0 kx + 3y – (k – 3) = 0 12x + ky – k = 0 kx + 3y – (k – 3) = 0 Comparing with a1x + b1y + c1 = 0 ∴ a1 = k , b1 = 3 , c1 = – (k – 3) 12x + ky – k = 0 Comparing with a2x + b2y + c2 = 0 ∴ a2 = 12 , b2 = k , c2 = –k Since equation has infinite number of solutions So, 𝑎1/𝑎2 = 𝑏1/𝑏2 = 𝑐1/𝑐2 Putting values 𝑘/12 = 3/𝑘 = (−(𝑘 − 3))/(−𝑘) 𝑘/12 = 3/𝑘 = (𝑘 − 3)/𝑘 Solving 𝒌/𝟏𝟐 = 𝟑/𝒌 k2 =12 × 3 k2 = 36 k = ±√36 k = ± √62 k = ± 6 So, k = 6, – 6 Solving 𝟑/𝒌 = (𝒌 −𝟑)/𝒌 3k = k(k -3) 3k = k2 – 3k 0 = k2 – 3k – 3k k2 – 3k – 3k = 0 k2 – 6k = 0 k(k – 6) = 0 So, k = 0 , k = 6 Since k = 6 satisfies both equations. Hence, k = 6 is the answer