Find the value of k for which each of the following system of equations have infinitely many solutions :
Find the value of k for which each of the following system of equations have infinitely many solutions :
$2 x+3 y=k$
$(k-1) x+(k+2) y=3 k$
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
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