Find the value of k for which each of the following system of equations have infinitely many solutions :

Solution:

GIVEN:

$k x+3 y=2 k+1$

$2(k+1) x+9 y=7 k+1$

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

For infinitely many solution

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

Here,

$\frac{k}{2(k+1)}=\frac{3}{9}=\frac{2 k+1}{7 k+1}$

Consider the following relation to find k

$\frac{k}{2(k+1)}=\frac{3}{9}$

$9 k=6(k+1)$

$9 k-6 k-6=0$

$3 k=6$

$k=2$

Now consider the following

$\frac{3}{9}=\frac{2 k+1}{7 k+1}$

$3(7 k+1)=9(2 k+1)$

$21 k+3=18 k+9$

$21 k-18 k=9-3$

$3 k=6$

$k=2$

Hence for $k=2$ the system of equation have infinitely many solutions