Find the value of k for the system of equations having infinitely many

1.	Find the value of k for the system of equations having infinitely many solution:2x + 3y = 2 (k+2)x + (2k+1)y = 2(k-1)
Answer» The given system of equations is: 2x + 3y – 2= 0 (k+2)x + (2k+1)y – 2(k-1) = 0 The above equations are of the form a₁ x + b₁ y − c₁ = 0 a₂ x + b₂ y − c₂ = 0 Here, a₁ = 2, b₁ = 3, c₁ = -5 a₂ = (k+2), b₂ = (2k+1), c₂ = -2(k-1) So according to the question, For unique solution, the condition is \(\frac{a_1 }{ a_2} = \frac{b_1 }{ b_2} = \frac{c_1 }{ c_2}\) \(\frac{2}{ (k+2)} = \frac{3}{ (2k+1)} = \frac{−2}{ −2(k−1)}\) 2/ (k+2) = 3/ (2k+1) and 3(2k+1) = 22(k−1) ⇒2(2k + 1) = 3(k + 2) and 3(k − 1) = (2k + 1) ⇒ 4k + 2 = 3k + 6 and 3k − 3 = 2k + 1 ⇒ k = 4 and k = 4 Hence, the given system of equations will have infinitely many solutions, if k = 4.

Find the value of k for the system of equations having infinitely many solution:2x + 3y = 2 (k+2)x + (2k+1)y = 2(k-1)

Answer»

The given system of equations is:

2x + 3y – 2= 0

(k+2)x + (2k+1)y – 2(k-1) = 0

The above equations are of the form

a₁ x + b₁ y − c₁ = 0

a₂ x + b₂ y − c₂ = 0

Here, a₁ = 2, b₁ = 3, c₁ = -5

a₂ = (k+2), b₂ = (2k+1), c₂ = -2(k-1)

So according to the question,

For unique solution, the condition is

\(\frac{a_1 }{ a_2} = \frac{b_1 }{ b_2} = \frac{c_1 }{ c_2}\)

\(\frac{2}{ (k+2)} = \frac{3}{ (2k+1)} = \frac{−2}{ −2(k−1)}\)

2/ (k+2) = 3/ (2k+1) and 3(2k+1) = 22(k−1)

⇒2(2k + 1) = 3(k + 2) and 3(k − 1) = (2k + 1)

⇒ 4k + 2 = 3k + 6 and 3k − 3 = 2k + 1

⇒ k = 4 and k = 4

Hence, the given system of equations will have infinitely many solutions, if k = 4.