Question

Find the value of k for which the following system of linear equations has an infinite number of solutions:

kx + 3y = (2k + 1), 2(k + 1)x + 9y = (7k + 1)

Answer 1

The given system of equations:

kx + 3y = (2k + 1)

⇒ kx + 3y – (2k + 1) = 0   ...(i)

And 2(k + 1)x + 9y = (7k + 1)

⇒ 2(k + 1)x + 9y – (7k + 1) = 0  ...(ii)

These equations are of the following form:

a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0

where, a₁ = k, b₁ = 3, c₁ = –(2k + 1) and a₂ = 2(k + 1), b₂ = 9, c₂ = –(7k + 1)

For an infinite number of solutions, we must have:

`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)`

i.e., `k/(2(k + 1)) = 3/9 = (-(2k + 1))/(-(7k + 1))`

⇒ `k/(2(k + 1)) = 1/3 = ((2k + 1))/((7k + 1))`

Now, we have the following three cases:

Case I:

`k/(2(k + 1)) = 1/3`

⇒ 2(k + 1) = 3k

⇒ 2k + 2 = 3k

⇒ k = 2

Case II:

`1/3 = ((2k + 1))/((7k + 1))`

⇒ (7k + 1) = 6k + 3

⇒ k = 2

Case III:

`k/(2(k + 1)) = ((2k + 1))/((7k + 1))`

⇒ k(7k + 1) = (2k + 1) × 2(k + 1)

⇒ 7k² + k = (2k + 1) (2k + 2)

⇒ 7k² + k = 4k² + 4k + 2k + 2

⇒ 3k² – 5k – 2 = 0

⇒ 3k² – 6k + k – 2 = 0

⇒ 3k(k – 2) + 1(k – 2) = 0

⇒ (3k + 1) (k – 2) = 0

⇒ k = 2 or k = `(-1)/3`

Hence, the given system of equations has an infinite number of solutions when k is equal to 2.