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

$2x\ â€“\ 3y\ =\ 7$

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

Given: The given equation are $2x\ –\ 3y\ =\ 7$ ; $(k\ +\ 2)x\ –\ (2k\ +\ 1)y\ =\ 3(2k\ -\ 1)$

To do: â€ŠFind the value of $k$ for which the following system of equations having infinitely many solutions.

Solution:

The given system of equation is:

$2x\ –\ 3y\ =\ 7$

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

The system of equation is of the form $a_{1} x+b_{1} y=c_{1}\ and\ a_{2} x+b_{2} y=c_{2}$

Here, $a_1 = 2, b_1=-3, c_1=-7 \ and \ a_2=k+2, b_2=2k+1, c_2=-3(2k-1) $

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{2}{k+2} =\frac{-3}{-(2k+1)} =\frac{-7}{-3(2k-1)} $

Now , $\frac{2}{k+2} =\frac{3}{2k+1}$ and $\frac{3}{2k+1} =\frac{7}{3(2k-1)} $

$2\times(2k+1) = 3(k+2)$ and $3(6k-3)=7\times(2k+1)$

$4k+2 = 3k+6$ and $18k-9=14k+7$

$4k-3k=6-2$ and $-14k+18k=9+7$

$k=4$ and $4k=16k$

$k=4$ and $k=\frac{16}{4} = 4$

Hence, the system of equations having infinitely many solutions if $k = 4$

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