Find the value of $k$ for which the following system of equations has no solution:
$3x\ -\ 4y\ +\ 7=\ 0$$kx\ +\ 3y\ -\ 5=\ 0$

Given: 

The given system of equations is:

$3x\ -\ 4y\ +\ 7=\ 0$

$kx\ +\ 3y\ -\ 5=\ 0$

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,

$3x\ -\ 4y\ +\ 7=\ 0$

$kx\ +\ 3y\ -\ 5=\ 0$

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,

$a_1=3, b_1=-4, c_1=7$ and $a_2=k, b_2=3, c_2=-5$

Therefore,

$\frac{3}{k}=\frac{-4}{3}≠\frac{7}{-5}$

$\frac{3}{k}=\frac{-4}{3}≠\frac{-7}{5}$

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

$3\times3=k\times(-4)$

$-4k=9$

$k=\frac{9}{-4}$

$k=-\frac{9}{4}$

The value of $k$ for which the given system of equations has no solution is $-\frac{9}{4}$. 

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Updated on: 10-Oct-2022

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