Question

answer it quickly please☝️​

Answer 1

Yo bro

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.