Question

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2. Find he value of k for which the system of equations x – 2y = 3 and 3x + ky = 1 has a unique

solution.

Answer 1

Hope it helps ...........

Answer 2

Given Equation is x - 2y - 3 = 0 ----- (1)

Given Equation is 3x + ky - 1 = 0  ------ (2)

On comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.

Here we have,

a1 = 1, b1 = -2, c1 = -3.

a2 = 3, b2 = k, c2 = -1

Given that the Equation has a unique solution.

$= \ \textgreater \ \frac{a1}{a2} \neq \frac{b1}{b2}$

$= \ \textgreater \ \frac{1}{3} \neq \frac{-2}{k}$

$= \ \textgreater \ k \neq -6$



The given system of linear equations has a unique solution for real values of k other than -6.



Hope this helps!