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

Solution:

GIVEN:

$2 x+3 y=k$

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

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 

Here,

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

Consider the following to find out k

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

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

$2 k+4=3 k-3$

$3 k-2 k=4+3$

$k=7$

Now again consider the following relation 

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

$3(3 k)=k(k+2)$

$9 k=k^{2}+2 k$

$k^{2}-7 k=0$

$k(k-7)=0$

$k=0$ or $(k-7)$

So the common solution is 7

Hence for $k=7$ the system of equation have infinitely many solutions