Question

find the value of k for which the pair
of linear equations 4x -3y-(k - 2) = 0 and
8x- 6y-K=0 has infinately many
solution​

Answer 1

Here,

$4x -3y-(k - 2) = 0$

$8x- 6y-k=0$

• It is of the form →

$\rm{}a_1x + b_1y + c_1 = 0 \\ \rm a_2x + b_2y + c_2 = 0$

From above,

$a_1 = 4, \:b_1 = - 3, \:c_1 = - (k - 2) \\ a_2 = 8, \:b_2 = - 6, \:c_2 = - k \: \: \: \: \: \: \: \: \: \:$

Now,

For infinitely many solutions,

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

So,

$= > \frac{4}{8} = \frac{ \cancel- 3}{ \cancel-6} = \frac{\cancel-( k- 2)}{ \cancel-k} \\ \\ = > \frac{1}{2} = \frac{1}{2} = \frac{k - 2}{k} \\ \\ \\ \huge = > \frac{1}{2} = \frac{ k- 2}{k} \\ \\ \\ \huge = > k = 2( k- 2) \\ \\ \\ \huge = > k = 2k - 4 \\ \\ \\ \huge = > 4 = 2k - k \\ \\ \\ \huge = > 4 = k \\ \\ \\ \huge = > k = 4$