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 = 7 (k+2)x – (2k+1)y = 3(2k-1)
Answer» The given system of equations is: 2x – 3y – 7 = 0 (k+2)x – (2k+1)y – 3(2k-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 = -7 a₂ = (k+2), b₂ = -(2k+1), c₂ = -3(2k-1) So according to the question, For unique solution, the condition is \(\frac{a_1 }{ a_2} = \frac{b_1 }{ b_2} = \frac{c_1}{ c_2}\) \(\frac{2}{(k+2)}\) = \(\frac{−3}{ −(2k+1)}\) = \(\frac{−7}{−3(2k−1)}\) \(\frac{2}{(k+2)}\) = \(\frac{−3}{ −(2k+1)}\) and \(\frac{−3}{ −(2k+1)}\) = \(\frac{−7}{−3(2k−1)}\) ⇒ 2(2k+1) = 3(k+2) and 3 x 3(2k−1) = 7(2k+1) ⇒ 4k + 2 = 3k + 6 and 18k – 9 = 14k + 7 ⇒ k = 4 and 4k = 16 ⇒ 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 = 7 (k+2)x – (2k+1)y = 3(2k-1)

Answer»

The given system of equations is:

2x – 3y – 7 = 0

(k+2)x – (2k+1)y – 3(2k-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 = -7

a₂ = (k+2), b₂ = -(2k+1), c₂ = -3(2k-1)

So according to the question,

For unique solution, the condition is

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

\(\frac{2}{(k+2)}\) = \(\frac{−3}{ −(2k+1)}\) = \(\frac{−7}{−3(2k−1)}\)

\(\frac{2}{(k+2)}\) = \(\frac{−3}{ −(2k+1)}\) and \(\frac{−3}{ −(2k+1)}\) = \(\frac{−7}{−3(2k−1)}\)

⇒ 2(2k+1) = 3(k+2) and 3 x 3(2k−1) = 7(2k+1)

⇒ 4k + 2 = 3k + 6 and 18k – 9 = 14k + 7

⇒ k = 4 and 4k = 16

⇒ k = 4

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