Find the value of $k$ for which the following system of equations has no solution:
$2x\ -\ ky\ +\ 3=\ 0$$3x\ +\ 2y\ -\ 1=\ 0$

Given: 

The given system of equations is:

$2x\ -\ ky\ +\ 3=\ 0$

$3x\ +\ 2y\ -\ 1=\ 0$

To do: 

We have to find the value of $k$ for which the given system of equations has no solution.

Solution:

The given system of equations is,

$2x\ -\ ky\ +\ 3=\ 0$

$3x\ +\ 2y\ -\ 1=\ 0$

The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.

The condition for which the above system of equations has no solution is

$\frac{a_{1}}{a_{2}} \ =\frac{b_{1}}{b_{2}} ≠ \frac{c_{1}}{c_{2}} \ $

Comparing the given system of equations with the standard form of equations, we have,

$a_1=2, b_1=-k, c_1=3$ and $a_2=3, b_2=2, c_2=-1$

Therefore,

$\frac{2}{3}=\frac{-k}{2}≠\frac{3}{-1}$

$\frac{2}{3}=\frac{-k}{2}≠-3$

$\frac{2}{3}=\frac{-k}{2}$

$2\times2=-k\times3$

$-3k=4$

$k=\frac{4}{-3}$

$k=-\frac{4}{3}$

The value of $k$ for which the given system of equations has no solution is $-\frac{4}{3}$. 

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Updated on: 10-Oct-2022

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