Question

26) please solve this question

Answer 1

Answer:

When k = (-1) the system of equations have no solution

Step-by-step explanation:

I hope you are asking Question 24. if not please specify the Question and I will do it again.

Now, we have,

(3k + 1)x + 3y - 2 = 0

(k² + 1)x + (k - 2)y - 5 = 0

It is of the form,

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Where,

a1 = (3k + 1)

a2 = (k² + 1)

b1 = 3

b2 = (k - 2)

c1 = (-2)

c2 = (-5)

Now, If the system of equation doesn't have a solution then,

(a1/a2) = (b1/b2) ≠ (c1/c2)

So,

(3k + 1)/(k² + 1) = 3/(k - 2) ≠ (-2)/(-5)

Lets first take the equals,

(3k + 1)/(k² + 1) = 3/(k - 2)

Cross multiplying,

(3k + 1)(k - 2) = 3(k² + 1)

3k² - 6k + k - 2 = 3k² + 3

3k² - 3k² - 5k - 2 = 3

-5k - 2 = 3

-5k = 3 + 2

-5k = 5

k = (5/-5)

k = (-5/5)

k = (-1)

Hence,

When k = (-1) the system of equations have no solution

Hope it helped you and believing you understood it...All the best