Question

S. The value of k for which the equations (3k + 1)x + 3y = 2: (k + 1)X + (k - 2)y = 5 has no solution, then k is equal to (a) 2 (b) 3 (C) 1 (d)-1. please ans with process​

Answer 1

Answer:

explai

Step-by-step explanation:

firstly you can put the value one by one and check when you put any number the one number is sutable for the answer

Answer 2

Step-by-step explanation:

bullet(3k+1)x+3y-2

k2+1)x+(k-2)y-5------2

on comparing the eq 1 nd 2....

a1=3k+1 , b1=3 , c1=-2

a2=k^ ^ 2+1,b2=k-2 , c2=-5

the equation have no solution.

SO,

a1/a2=b1/b2\#c1/c2

3k + 1 / (k ^ 2) + 1 = 3 / k - 2 .

3k^ ^ 2-6k+k-2=3k^ ^ 2+3

k - 2 = 3

> - 5k = 5

k = - 1

I hope it help you