Question

For what value of k the following pair has infinite number of solutions (k-3)x+3y=k,k(x+y)=12.​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

• For what value of k the following pair has infinite number of solutions (k-3)x + 3y = k, k(x+y) = 12.

Answer

Given :-

Two lines,

• (k - 3)x + 3y = k and kx + ky = 12

To Find :-

• Value of 'k' for which lines has infinitely many solutions.

Understanding the concept :-

$\sf \: Let \: consider \: lines \: a_1x + b_1y + c_1 = 0 \: and \: a_2x + b_2y + c_2 = 0$

then,

• two lines have infinitely many solutions iff

$\bf \:\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

$\large\underline{\bold{Solution - }}$

Given that

• two lines (k - 3)x + 3y = k and kx + ky = 12 have infinitely many solutions

Now,

We know,

• Two lines have infinitely many solutions, iff

$\bf \:\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

Here,

• a₁ = k - 3

• a₂ = k

• b₁ = 3

• b₂ = k

• c₁ = k

• c₂ = 12

Now,

• On substituting the values, we get

$\rm :\implies\:\dfrac{k - 3}{k} = \dfrac{3}{k} = \dfrac{k}{12}$

• On taking first two members, we have

$\rm :\longmapsto\:\dfrac{k - 3}{k} = \dfrac{3}{k}$

$\rm :\longmapsto\: {k}^{2} - 3k = 3k$

$\rm :\longmapsto\: {k}^{2} - 6k = 0$

$\rm :\longmapsto\:k(k - 6) = 0$

$\bf\implies \: k \: = \: 0 \: \: \: \: or \: \: \: \: \: k \: = \: 6 - - - (1)$

Now,

• On taking last two members, we have

$\rm :\longmapsto\:\dfrac{3}{k} = \dfrac{k}{12}$

$\rm :\longmapsto\: {k}^{2} = 36$

$\rm :\longmapsto\:k = \pm \: 6$

$\bf\implies \: k \: = \: - 6 \: \: \: \: or \: \: \: \: \: k \: = \: 6 - - - (2)$

Hence,

• From equation (1) and equation (2), we concluded

$\bf\implies \: k \: = \: 6 \: \: \: \:$

Additional Information :-

Let us consider two linear equations

$\tt \: a_1x + b_1y + c_1 = 0 \: and \: a_2x + b_2y + c_2 =0$

then

• (1). System of equations have unique solution iff

$\bf \:\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$

• (2). System of equations have no solution iff

$\bf \:\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$