Question

Kx+3y-(k-3)=0

12x+ky-k=0 solved with substantial method​

Answer 1

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Answer 2

Answer:

k=6

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

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