Question

Find the value of k if the equations 3x + (k/3 +2)y=1 and kx +2ky =4 have infinite solutions

Answer 1

$\huge{ \underline{ \underline{ \green{ \sf{ Detailed \: Answer :}}}}}$

To find ; value of k

if the equations 3x + (k/3 +2)y=1 and kx +2ky =4 have infinite solutions.

Concept :

The pair of linear equations represented by these lines.

$a_{1}x \: + b_{1}y + c_{1} = 0$

$a_{2}x + b _{2}y + c_{2} = 0$

has infinetly many solutions if

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

${\blue {\huge{\underline{\mathbb{Answer:-}}}}}$

Given linear equations are :

3x + (k/3 +2)y=1 and kx +2ky =4

since they have infinetly many solutions;

therefore ,

$\frac{3}{k} = \frac{ \frac{k}{3} + 2}{2k} = \frac{1}{4}$

$\frac{3}{k} = \frac{ \frac{k + 6}{3} }{2k} = \frac{1}{4}$

Now equate :

$\frac{k + 6}{6k} = \frac{1}{4}$

$4k + 24 = 6k$

$2k = 24$

k = 12

${\purple{\boxed{\large{\bold{K =12 }}}}}$