1.

For what value of k, the following system of equations has a unique solution : 2x + 3y – 5 = 0, kx – 6y – 8 = 0 ? (a) k = –4 (b) k \(\ne\) – 4 (c) k \(\ne\) 4 (d) k = 4

Answer»

(b) k \(\ne\) – 4

In the given system of equations,

a1 = 2, b1 = 3, c1 = –5 

a2 = k, b2 = –6, c2 = – 8

For a unique solution.

\(\frac{a_1}{a_2}\)\(\ne\)\(\frac{b_1}{b_2}\) \(\Rightarrow\) \(\frac{2}{k}\) \(\ne\) \(\frac{3}{-6}\) \(\Rightarrow\) 3k \(\ne\) -12

\(\Rightarrow\) k\(\ne\) -4



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