Question

The pair of linear equations (3k+1)x+3y-5=0 and 2x-3y+5=0 have infinite solutions.Then find the value of k.

Answer 1

Given linear equations are
(3k+1)x+3y-5=0  ........ (1)
2x-3y+5=0    ......... (2)
We have to find the value of "k"
For this, add equation (1) and (2)
we get,
[ (3k + 1) x + 3y - 5 ] + [ 2x - 3y + 5 ] = 0
(3k + 1) x + 3y - 5 + 2x - 3y + 5 = 0
3kx + x + 3y - 5 + 2x - 3y + 5 = 0
On simplification, we get:
3kx + 3x = 0
Taking 3x common:
3x (k + 1) = 0
k + 1 = 0
k = -1
which is the required value of "k".
Hope it will help you. Thanks.

Answer 2

$\huge\bf\fbox\red{Answer:-}$

k = -1

Explanation:

Given linear equations are

(3k + 1)x + 3y - 5 = 0 --- (1)

2x - 3y + 5 = 0 --- (2)

We have to find the value of "k"

For this, add equation (1) and (2)

We get,

$[(3k+1)x+3y-5]+[2x-3y + 5] = 0 \\ (3k+1)x+3y-5+2x-3y + 5 = 0 \\ 3kx+x+3y-5+ 2x - 3y + 5 = 0$

On simplification, we get:

3kx + 3x = 0

Taking 3x common:

3x (k+ 1) = 0

k + 1 = 0

k = -1

Hence, -1 is the required value of "k".