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Find the value of K for which equation x – Ky = 2, 3x + 2y = 5 has unique solution.
Question

Find the value of K for which equation x – Ky = 2, 3x + 2y = 5 has unique solution.

A.

K23K \ne -\frac{2}{3} \quad​​

B.

K=23K = \frac{2}{3} \quad​​

C.

K23K \ne \frac{2}{3} \quad​​

D.

K=23 K = -\frac{2}{3} \\[12pt]​​

Correct option is A

Given equations:
xKy=2=>a1=1, b1=K3x+2y=5=>a2=3, b2=2x - Ky = 2 \quad \Rightarrow a_1 = 1,\ b_1 = -K \\3x + 2y = 5 \quad \Rightarrow a_2 = 3,\ b_2 = 2 \\[8pt]​​
For unique solution:
a1a2b1b2=>13K2\frac{a_1}{a_2} \ne \frac{b_1}{b_2} \Rightarrow \frac{1}{3} \ne \frac{-K}{2} \\​​
Cross-multiplying: 23K=>K232 \ne -3K \Rightarrow K \ne -\frac{2}{3} \\[10pt]​​
Correct answer is (a) K23K \ne -\frac{2}{3}​​​​

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