Question

from the pair of equations kx+3y= -7, 2x-6y=14 to have infinitely many solutions, find the value of k.

Answer 1

Answer:

Step-by-step explanation:

Kx+3y+7=0

2x-6y-14=0

Here,

a1=K,b1=3,c1=7

a2=2,b2=-6 , c3=-14

For infinitely many soln.

a1/a2=b1/b2=c1/c2

So,

K/2=3/-6=7/-14

K/2=-1/2

So,

K=-1

Answer 2

Answer:

Required value of k is (-1)

Step-by-step explanation:

Given equations are $kx + 3y = - 7.....(1)$and $2x - 6y = 14........(2)$

We are comparing equation (1) with $a_1x + b_1y = c_1$

So we get

$a_1 = k \\ b_1 = 3 \\ c_1 = - 7$

Again we are comparing equation (2) with

$a_2x + b_2y = c_2$ and get

$a_2 = 2 \\ b_2 = - 6 \\ c_2 = 14$

Now it is given that these two equations have infinite many solutions.

We know if two equations $a_1x + b_1y = c_1$and $a_2x + b_2y = c_2$

have infinite solution then

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

So,

$\frac{k}{2} = \frac{3}{ - 6} = \frac{ - 7}{14}$

Or,$\frac{k}{2} = - \frac{1}{2}$

By cross multiplication,

$2 \times k = - 1 \times 2$

$k = - \frac{2}{2} = - 1$

Required value of k is (-1)