Question

for what value of k the pair of equations kx -4y=3,6x-12y=9has infinite number of solutions

Answer 1

kx-4y=3......¡
6x-12y=9....¡¡

If it has infinite number of solutions then
a1/a2=b1/b2=c1/c2

a1/a2=b1/b2

=>k/6=4/12

=>12k=24

=>k=24/12

=>k=2

b1/b2=c1/c2

=>4/12=3/9

=>1/3=1/3

Therefore the value of k is 2

Answer 2

$\huge{\underline{\underline{\blue{\mathfrak{Answer :}}}}}$

$\huge{\underline{\bf{Given \: :}}}$

⟹ kx - 4y = 3

⟹ 6x - 12y = 9

______________________________________

$\huge{\underline{\bf{To \: Find \: :}}}$

Value of k.

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$\huge{\underline{\bf{Solution \: :}}}$

$\sf \rightarrow {\frac{a_{1}}{a_{2}}} = \sf{\frac{b_{1}}{b_{2}}} = \sf {\frac{c_{1}}{c_{2}}} \\ \\ \sf \rightarrow { \frac{k}{6} = \frac{ \cancel{ - 4}}{ \cancel{- 12}}} \\ \\ \sf \rightarrow { \frac{k}{6} = \frac{1}{3} } \\ \\ \sf \rightarrow {3k = 6} \\ \\ \sf \rightarrow {k = \frac{ \cancel6}{ \cancel3} } \\ \\ \sf \rightarrow { k = 2}$

$\rule{200}{2}$

$\huge{\underline{\bf{Verification\: :}}}$

$\sf{ \frac{ \cancel2}{\cancel6} = \frac{\cancel{ - 4}}{ \cancel{- 12}} = \frac{\cancel3}{\cancel9} } \\ \\ \sf{ \frac{1}{3} = \frac{1}{3} = \frac{1}{3} }$

★ Hence Verified