Question

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

2x + (k – 2)y = k, 6x + (2k – 1)y = (2k + 5)

Answer 1

The given system of equations:

2x + (k – 2)y = k

⇒ 2x + (k – 2)y – k = 0   ...(i)

And 6x + (2k – 1)y = (2k + 5)

⇒ 6x + (2k – 1)y – (2k + 5) = 0   ...(ii)

These equations are of the following form:

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

where, a₁ = 2, b₁ = (k – 2), c₁ = –k and a₂ = 6, b₂ = (2k – 1), c₂ = –(2k + 5)

For an infinite number of solutions, we must have:

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

`2/6 = ((k - 2))/((2k - 1)) = (-k)/(-(2k + 5))`

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

Now, we have the following three cases:

Case I:

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

⇒ (2k – 1) = 3(k – 2)

⇒ 2k – 1 = 3k – 6

⇒ k = 5

Case II:

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

⇒ (k – 2) (2k + 5) = k(2k – 1)

⇒ 2k² + 5k – 4k – 10 = 2k² – k

⇒ k + k = 10

⇒ 2k = 10

⇒ k = 5

Case III:

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

⇒ 2k + 5 = 3k

⇒ k = 5

Hence, the given system of equations has an infinite number of solutions when k is equal to 5.