Find the value of $k$ for which the following system of equations having infinitely many solution:
$2x\ +\ 3y\ –\ 5\ =\ 0$
$6x\ +\ ky\ –\ 15\ =\ 0$

Given: The system of equation is

$2x\ +\ 3y\ –\ 5\ =\ 0$; $6x\ +\ ky\ –\ 15\ =\ 0$

 

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:

$2x\ +\ 3y\ –\ 5\ =\ 0$

$6x\ +\ ky\ –\ 15\ =\ 0$

The given system of equation is in the form

$a_1x+b_1y+c_1=0$

$a_2x+b_2y+c_2=0$

Here, $a_1=2,b_1=3 ,c_1=-5 ; a_2=6,b_2=k,c_2=-15$

For unique solution we must have: 

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\frac{2}{6}=\frac{3}{k}=\frac{-5}{-15}$

$\frac{1}{3}=\frac{3}{k}=\frac{1}{3}$

$\frac{1}{3}=\frac{3}{k}$ and$\frac{3}{k}=\frac{1}{3}$

$k=9$ and $k=9$

K=9 satisfies both the conditions

Therefore, given system of the equation has infinitely many solutions at $k=9$

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Updated on: 10-Oct-2022

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