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Find the value of $k$ for which the system
$kx+2y=5$
$3x+y=1$
has no solution.
Given:
The given system of equations is:
$kx+2y=5$
$3x+y=1$
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 can be written as:
$kx+2y-5=0$
$3x+y-1=0$
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,
$a_1=k, b_1=2, c_1=-5$ and $a_2=3, b_2=1, c_2=-1$
Therefore,
$\frac{k}{3}=\frac{2}{1}≠\frac{-5}{-1}$
$\frac{k}{3}=2≠5$
$\frac{k}{3}=2$
$k=3\times2$
$k=6$
The value of $k$ for which the given system of equations has no solution is $6$.