Question

If kx+3y-(k-3)=0,12x+ky-k=0 for what values of k will the following pair of linear equations have infinitely many solutions?

Answer 1

Answer:

k=6

Step-by-step explanation:

The given equations are:

$kx+3y-(k-3)=0$ and $12x+ky-k=0$

Therefore, $a_{1}=k$, $a_{2}=12$, $b_{1}=3$, $b_{2}=k$, $c_{1}=-(k-3)$, $c_{2}=-k$

Since, the equations has infinitely many solutions, therefore

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

$\frac{k}{12}=\frac{3}{k}=\frac{-(k-3)}{-k}$

Taking second and third equalities,

$3=k-3$

$k=6$

Answer 2

Answer: Hope it help u all........

Step-by-step explanation:

Thank u.........