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

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

$6x\ +\ (2k\ -\ 1)y\ =\ 2k\ +\ 5$

**Given**: The system of equation is $2x\ +\ (k\ -\ 2)y\ =\ k$ ; $6x\ +\ (2k\ -\ 1)y\ =\ 2k\ +\ 5$

**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:

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

$6x\ +\ (2k\ -\ 1)y\ =\ 2k\ +\ 5$

The given system of equation is in the form

$a_1x+b_1y+c_1=0$

$a_2x+b_2y+c_2=0$

Here, $a_1=2.b_1=k-2 ,c_1=-k ; a_2=6.b_2=2k-1,c_2=-(2k+5)$

For unique solution we must have:

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

$\frac{2}{6}=\frac{k-2}{2k-1}=\frac{-k}{-(2k+5)}$

$\frac{1}{3}=\frac{k-2}{2k-1}=\frac{k}{2k+5}$

$\frac{1}{3}=\frac{k-2}{2k-1}$ and $\frac{k-2}{2k-1}=\frac{k}{2k+5}$

$3(k-2)=2k-1$ and $(2k+5)(k-2)=k(2k-1)$

$3k-6=2k-1$and $2k^2+5k-4k-10=2k62-k$

$3k-2k=5$ and $k-10=-k$

$k=5$and $2k=10=k=5$

K=5 satisfies both the conditions

**Therefore, given system of the equation has infinitely many solutions at $k=5$**