Question

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

5x + 2y = 2k, 2(k + 1)x + ky = (3k + 4)

Answer 1

The given system of equations:

5x + 2y = 2k

⇒ 5x + 2y – 2k = 0   ...(i)

And 2(k + 1)x + ky = (3k + 4)

⇒ 2(k + 1)x + ky – (3k + 4) = 0   ...(ii)

These equations are of the following form:

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

where, a₁ = 5, b₁ = 2, c₁ = –2k and a₂ = 2(k + 1), b₂ = k, c₂ = –(3k + 4)

For an infinite number of solutions, we must have:

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

`5/(2(k + 1)) = 2/k = (-2k)/(-(3k + 4))`

⇒ `5/(2(k + 1)) = 2/k = (2k)/(3k + 4)`

Now, we have the following three cases:

Case I:

`5/(2(k + 1)) = 2/k`

⇒ 2 × 2(k + 1) = 5k

⇒ 4(k + 1) = 5k

⇒ 4k + 4 = 5k

⇒ k = 4

Case II:

`2/k = (2k)/(3k + 4)`

⇒ 2k² = 2 × (3k + 4)

⇒ 2k² = 6k + 8

⇒ 2k² – 6k – 8 = 0

⇒ 2(k² – 3k – 4) = 0

⇒ k² – 4k + k – 4 = 0

⇒ k(k – 4) + 1(k – 4) = 0

⇒ (k + 1) (k – 4) = 0

⇒ (k + 1) = 0 or (k – 4) = 0

⇒ k = –1 or k = 4

Case III:

`5/(2(k + 1)) = (2k)/((3k + 4))`

⇒ 15k + 20 = 4k² + 4k

⇒ 4k² – 11k – 20 = 0

⇒ 4k² – 16k + 5k – 20 = 0

⇒ 4k(k – 4) + 5(k – 4) = 0

⇒ (k – 4) (4k + 5) = 0

⇒ k = 4 or k = `(-5)/4`

Hence, the given system of equations has an infinite number of solutions when k is equal to 4.