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Find the values of $ k $ for which the system
$2 x+k y=1$
$ 3 x-5 y=7 $
will have a unique solution.
Given:
The given system of equations is:
\(2 x+k y=1\)
\( 3 x-5 y=7 \)
To do:
We have to find the value of $k$ for which the given system of equations will have a unique solution.
Solution:
The given system of equations can be written as:
$2x+ky-1=0$
$3x-5y-7=0$
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 a unique solution is:
$\frac{a_{1}}{a_{2}} \ ≠ \frac{b_{1}}{b_{2}}$
Comparing the given system of equations with the standard form of equations, we have,
$a_1=2, b_1=k, c_1=-1$ and $a_2=3, b_2=-5, c_2=-7$
Therefore,
$\frac{2}{3}≠\frac{k}{-5}$
$k≠ \frac{-5\times2}{3}$
$k≠ \frac{-10}{3}$
The value of $k$ for which the given system of equations has a unique solution is $k≠ \frac{-10}{3}$.