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Find the value of $k$ for which the following system of equations having infinitely many solution:
$kx\ –\ 2y\ +\ 6\ =\ 0$
$4x\ –\ 3y\ +\ 9\ =\ 0$
$kx\ –\ 2y\ +\ 6\ =\ 0$
$4x\ –\ 3y\ +\ 9\ =\ 0$
Given: The given equation are $kx\ –\ 2y\ +\ 6\ =\ 0$ ;$4x\ –\ 3y\ +\ 9\ =\ 0$
To do:  Find the value of $k$ for which the following system of equations having infinitely many solutions.
Solution:
The given system of equation is:
$kx\ –\ 2y\ +\ 6\ =\ 0$
$4x\ –\ 3y\ +\ 9\ =\ 0$
The system of equation is of the form $a_{1} x+b_{1} y=c_{1}\ and\ a_{2} x+b_{2} y=c_{2}$
For the infinitely many solutions there is a condition
$\frac{a_{1}}{a_{2}} \ =\frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}} \ $
$\frac{k}{4} =\frac{2}{3} =\frac{6}{9} \ $
Now , $\frac{k}{4} =\frac{2}{3}$
$\Rightarrow k\times3 = 4\times2$
$\Rightarrow 3k = 8$
$\Rightarrow k = \frac{3}{8}$
Hence, the system of equations having infinitely many solutions if $k =\frac{3}{8}$
 
Updated on: 10-Oct-2022
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