Question

tell pls fast !!?!!!!!???? Ans 23​

Answer 1

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Given:

• Equations=> 2x - ky + 3 = 0, 3x + 2y - 1 = 0

To Find:

• Value of k for which no solution exists.

Concept Used:

• For no solution to exist in a system of linear equations, $\dfrac{x_1}{x_2} \neq \dfrac{y_1}{y_2}$

Solution:

$x_1 = 2, x_2 = 3, y_1=-k, y_2 = 2.$

So, for no solution to exist in a system of linear equations, $\\ \dfrac{x_1}{x_2} \neq \dfrac{y_1}{y_2}$

$\\ \implies \dfrac{2}{3} \neq \dfrac{-k}{2}$

$\\ \implies \dfrac{4}{3} \neq \dfrac{-k}{1}$

$\\ \implies k \neq \dfrac{-4}{3}$

So, all values of k except $\dfrac{-4}{3}$ makes the system with no solution.

Hope it helps!!!!

Answer 2

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