Question

Find the value of K for which the given system of equation has infinite many solution. Kx+3y=k.3
12x+ky=k.

Answer 1

Answer :

The value of is 6 and -6

Given :

The pair of equations :

• kx + 3y = k.3
• 12x + ky = k
• The given pair of equations has infinitely many solutions.

To Find :

• The value of k

Concept to be used :

$\sf \bullet \: \: \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}, \: there \: there \: infinitely \: many \: solutions.$

Solution :

Consider the equations as :

$\sf \implies kx + 3y = 3k ..........(1)$

$\sf \implies 12x + ky = k ...........(2)$

Since (1) and (2) has infinitely many solutions so :

$\sf \dfrac{k}{12} = \dfrac{3}{k} = \dfrac{3k}{k} \\\\ \sf \implies \dfrac{k}{12} = \dfrac{3}{k} = 3$

Now taking a pair we have :

$\sf \implies \dfrac{k}{12}=\dfrac{3}{k}\\\\ \sf \implies k^{2} = 36 \\\\ \sf \implies k =\pm\sqrt{36} \\\\ \sf \implies k = \pm 6$

Thus , either k = 6 or k = -6

Answer 2

Answer:

Given

Kx + 3y = K - 3

Comparing with $a_{1x} + b_{1y} + c_{1} = 0$

$a_{1} = K ; b_{1} = 3 ; c_{1} = K-3$

12x + Ky - K = 0

Comparing with $a_{2} + b_{2} + c_{2} = 0$

$a_{2} = 12 ; b_{2} = K; c_{2} = -K$

Since equation has infinite number of solutions

$So, \frac{a_{1} }{a_{2} } = \frac{b_{1} }{b_{2} } = \frac{c_{1} }{c_{2} } \\\\\frac{K}{12} = \frac{3}{K} =\frac{K-3}{K} \\\\\frac{K}{12} =\frac{3}{K}$

$K^{2}$ =12×3

 $K^{2}$=36

k=±6

$\frac{3}{K} = \frac{K-3}{K}$

​ 3K=K (K−3)

3K = $K^{2}$   −3k  

$K^{2}$ −3K−3K=0

 $K^{2}$ −6K=0

K (K−6)=0

K=0,6

Therefore, K=6 satisfies both equations

Hence, K=6