The value of k for which the system of equations (3k + 1) x + 3y – 2

1.	The value of k for which the system of equations (3k + 1) x + 3y – 2 = 0 and (k2 + 1) x + (k – 2) y – 5 = 0 has no solution is…..............A) 1 B) 2 C) -1 D) -2
Answer» Correct option is (C) -1 For no solution, we have \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\) \(\Rightarrow\) \(\frac{3k+1}{k^2+1}=\frac{3}{k-2}\neq\frac{-2}{-5}\) \(\Rightarrow\) \(\frac{3k+1}{k^2+1}=\frac{3}{k-2}\) \(\Rightarrow\) \((3k+1)(k-2)=3(k^2+1)\) (By cross multiplication) \(\Rightarrow\) \(3k^2+k-6k-2\) \(=3k^2+3\) \(\Rightarrow\) -5k - 2 = 3 \(\Rightarrow\) -5k = 3+2 = 5 \(\Rightarrow\) k = \(\frac5{-5}\) = - 1 Correct option is C) -1

The value of k for which the system of equations (3k + 1) x + 3y – 2 = 0 and (k2 + 1) x + (k – 2) y – 5 = 0 has no solution is…..............A) 1 B) 2 C) -1 D) -2

Answer»

Correct option is (C) -1

For no solution, we have

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

\(\Rightarrow\) \(\frac{3k+1}{k^2+1}=\frac{3}{k-2}\neq\frac{-2}{-5}\)

\(\Rightarrow\) \(\frac{3k+1}{k^2+1}=\frac{3}{k-2}\)

\(\Rightarrow\) \((3k+1)(k-2)=3(k^2+1)\) (By cross multiplication)

\(\Rightarrow\) \(3k^2+k-6k-2\) \(=3k^2+3\)

\(\Rightarrow\) -5k - 2 = 3

\(\Rightarrow\) -5k = 3+2 = 5

\(\Rightarrow\) k = \(\frac5{-5}\) = - 1

Correct option is C) -1