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Find the number of values of $k$ for which the system of equations $x+y=2,\ kx+y=4,\ x+ky=5$ has only one solution.
Given: The system of equations $x+y=2,\ kx+y=4,\ x+ky=5$ has only one solution.
To do: To find the number of values of $k$.
Solution:
$x+y=2\ ....( i)$
$kx+y=4\ ....( ii)$
$x+ky=5\ ....( iii)$
On adding $( ii)$ and $( iii)$, we get
$kx+y+x+ky=4+5$
$( k+1)( x+y)=9$
But $x+y=2$
$\Rightarrow k+1=\frac{9}{2}=4.5$
$\Rightarrow k=4.5-1=3.5$
$\therefore$ For the given system of equation to have atleast one solution, $k=3.5$
Thus, $k$ has only $1$ value.
Updated on: 10-Oct-2022
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