Question

For what value of k, the pair of linear equations 3x+2y=4 and 6x+ky=8 will have infinite number of solutions. For what value of k, the pair of linear equations 3x+2y=4 and 6x+ky=8 will have infinite number of solutions.
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Answer 1

Question:—

For what value of k, the pair of linear equations 3x+2y=4 and 6x+ky=8 will have infinite number of solutions ?

Solution:—

Here ,

$\bold{a_1 = 3 , \: \: \: \: b_1= 2 , \: \: \: \: \: c_1 = 4}$

$\bold{a_2 = 6 , \: \: \: \: \: b_2 = k , \: \: \: \: \: \: c_2 = 8}$

★ We know that , any two liner equations

$a_1x + b_1y + c_1 = 0$

$and \: \: \: a_2x + b_2y + c_2 = 0 \: \: will \: \: have \: \: infinite \: \: \: number \: \: of \: \: solutions$

$If \: \: \: \: \boxed{\bold{ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} }}$

So,

$\large \color{blue}\bold { \frac{3}{6} = \frac{2}{k} = \frac{4}{8} }$

$\implies \: \large \color{blue}\bold { \frac{3}{6} = \frac{2}{k} \: \: \: \color{red}or \: \: \: \color{blue}{\frac{2}{k} = \frac{4}{8} }}$

$\implies \: \large \color{blue}\bold { \frac{1}{2} = \frac{2}{k} \: \: \: \color{red}{ or} \: \: \: \color{blue}{\frac{2}{k} = \frac{1}{2} }}$

$\implies \: \large \color{blue}\bold {k = 4 \: \: \: \color{red}or \color{blue}{\: \: \: k = 4}}$

$\huge{Hence , }$

$\huge \underbrace {\bold \color{orange} {k = 4}}$