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

Solution:

GIVEN:

$2 x+3 y=7$

$(k+1) x+(2 k-1) y=4 k+1$

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

$\frac{2}{(k+1)}=\frac{3}{2(k-1)}=\frac{7}{4 k+1}$

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

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

$4 k-2=3 k+3$

$4 k-3 k=2+3$

$k=5$

Now again consider the following to find k

$\frac{3}{2(k-1)}=\frac{7}{4 k+1}$

$3(4 k+1)=7 \times 2(k-1)$

$12 k+3=14 k-14$

$14+3=14 k-12 k$

$k=5$

Hence for $k=5$ the system of equation have infinitely many solutions