Question

For what value of K. the system of equation has infinite many solution ? kx+3y - ( k-3) = 0 ,
1 2x + ky - k = 0​

Answer 1

Answer:

Step-by-step explanation:

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