For what value of k, the system of equations kx – y = 2, 6x – 2y �

1.	For what value of k, the system of equations kx – y = 2, 6x – 2y – 3 has infinitely many solutions ? A) k = 2 B) k = 3C) k = 1 D) Does not exists
Answer» Correct option is (D) Does not exists Given system of equations is kx – y = 2 \(\Rightarrow\) kx - y - 2 = 0 _________(1) 6x – 2y – 3 = 0 _________(2) By comparing given system with standard form of system, we get \(a_1=k,b_1=-1,c_1=-2\) and \(a_2=6,b_2=-2,c_2=-3\) \(\therefore\) \(\frac{a_1}{a_2}=\frac k6,\) \(\frac{b_1}{b_2}=\frac{-1}{-2}=\frac12\) \(\frac{c_1}{c_2}=\frac{-2}{-3}=\frac23\) \(\because\) \(\frac12\neq\frac23\) \(\therefore\) \(\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\) But condition for infinitely many solutions is \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) which is not possible. Hence, for any value of k, the given system never give infinitely many solutions. Hence, the value of k does not exists. Correct option is D) Does not exists

For what value of k, the system of equations kx – y = 2, 6x – 2y – 3 has infinitely many solutions ? A) k = 2 B) k = 3C) k = 1 D) Does not exists

Answer»

Correct option is (D) Does not exists

Given system of equations is

kx – y = 2

\(\Rightarrow\) kx - y - 2 = 0 _________(1)

6x – 2y – 3 = 0 _________(2)

By comparing given system with standard form of system, we get

\(a_1=k,b_1=-1,c_1=-2\)

and \(a_2=6,b_2=-2,c_2=-3\)

\(\therefore\) \(\frac{a_1}{a_2}=\frac k6,\)

\(\frac{b_1}{b_2}=\frac{-1}{-2}=\frac12\)

\(\frac{c_1}{c_2}=\frac{-2}{-3}=\frac23\)

\(\because\) \(\frac12\neq\frac23\)

\(\therefore\) \(\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\)

But condition for infinitely many solutions is

\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) which is not possible.

Hence, for any value of k, the given system never give infinitely many solutions.

Hence, the value of k does not exists.

Correct option is D) Does not exists