Find the value of $k$ for which the following system of equations having infinitely many solutions:

$2x\ +\ 3y\ =\ 2$
$(k\ +\ 2)x\ +\ (2k\ +\ 1)y\ =\ 2(k\ -\ 1)$

Given: The system of equation is

$2x\ +\ 3y\ =\ 2$; $(k\ +\ 2)x\ +\ (2k\ +\ 1)y\ =\ 2(k\ -\ 1)$

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:

$2x\ +\ 3y\ =\ 2$

$(k\ +\ 2)x\ +\ (2k\ +\ 1)y\ =\ 2(k\ -\ 1)$

The given system of equation is in the form

$a_1x+b_1y+c_1=0$

$a_2x+b_2y+c_2=0$

Here, $a_1=2,b_1=3 ,c_1=-2 ; a_2=(k+2),b_2=2k+1,c_2=-2(k-1)$

For unique solution we must have: 

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

$\frac{2}{k+2}=\frac{3}{2k+1}=\frac{-2}{-2(k-1)}$

$\frac{2}{k+2}=\frac{3}{2k+1}=\frac{2}{2k-2}$

$\frac{2}{k+2}=\frac{3}{2k+1}$ and$\frac{3}{2k+1}=\frac{2}{2k-2}$

$2(2k+1)=3(k+2)$ and $3(2k-2)=2(2k+1)$

$4k+2=3k+6$ and $6k-6=4k+2$

$4k-3k=-2+6$ and $6k-4k=+6+2$

$k=4$ and $2k=8=k=4$ 

K=4 satisfies both the conditions

Therefore, given system of the equation has infinitely many solutions at $k=4$

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Updated on: 10-Oct-2022

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