Find the value of $k$ for which the following system of equations has a unique solution:
$kx\ +\ 2y\ =\ 5$
$3x\ +\ y\ =\ 1$

 Given: The system of equation is

$kx\ +\ 2y\ =\ 5$; $3x\ +\ y\ =\ 1$

To do: Find the value of $k$ for which the system of the equation has infinitely many solutions.

Solution: 

The given system of the equation can be written as:

$kx\ +\ 2y\ =\ 5$

$3x\ +\ y\ =\ 1$

The given system of equation is in the form

$a_1x+b_1y+c_1=0$

$a_2x+b_2y+c_2=0$

Here, $a_1=k,b_1=2 ,c_1=5 ; a_2=3,b_2=1,c_2=1$

For unique solution we must have: 

$\frac{a_1}{a_2} $ not equal to $\frac{b_1}{b_2}$

$\frac{k}{3}$ not equal to $\frac{2}{1}$

$\frac{k}{3}$ not equal to $2$

$k$ not equal to $3\times2$

$k$ not equal to $6$

So, the given system of equation will have a unique solution for all real values except $k=6$

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Updated on: 10-Oct-2022

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