Question

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

2x + 3y = 7, (k – 1)x + (k + 2)y = 3k

Answer 1

The given system of equations:

2x + 3y = 7,

⇒ 2x + 3y – 7 = 0   ...(i)

And (k – 1)x + (k + 2)y = 3k

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

These equations are of the following form:

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

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

For an infinite number of solutions, we must have:

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

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

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

Now, we have the following three cases:

Case I:

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

⇒ 2(k + 2) = 3(k – 1)

⇒ 2k + 4 = 3k – 3

⇒ k = 7

Case II:

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

⇒ 7(k + 2) = 9k

⇒ 7k + 14 = 9k

⇒ 2k = 14

⇒ k = 7

Case III:

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

⇒ 7k – 7 = 6k

⇒ k = 7

Hence, the given system of equations has an infinite number of solutions when k is equal to 7.