Find the value of $k$ for which the following system of equations has a unique solution:
$4x\ +\ ky\ +\ 8\ =\ 0$
$2x\ +\ 2y\ +\ 2\ =\ 0$

 Given: The system of equation is

$4x\ +\ ky\ +\ 8\ =\ 0$ ; $2x\ +\ 2y\ +\ 2\ =\ 0$

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:

$4x\ +\ ky\ +\ 8\ =\ 0$

$2x\ +\ 2y\ +\ 2\ =\ 0$

The given system of equation is in the form

$a_1x+b_1y+c_1=0$

$a_2x+b_2y+c_2=0$

Here, $a_1=4,b_1=k ,c_1=8 ; a_2=2,b_2=2,c_2=2$

For unique solution we must have: 

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

$\frac{4}{2}$ not equal to $\frac{k}{2}$

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

$k$ not equal to 4

So, the given system of equation will have a unique solution for all real values except $k=4$

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Updated on: 10-Oct-2022

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