# Find the values of k?

Find the values of k such that the following system has a) no solution, or b) a unique one.

x+2y+2z=1

y+kz=1

-x+y+kz=k

### 2 Answers

- GeronimoLv 77 years agoFavorite Answer
x + 2y + 2z = 1 ... equation 1 (given)

- x + y + kz = k ... equation 3 (given)

——————————<ADD>

3y + (2 + k)z = 1 + k ... equation A

y + kz = 1 ... equation 2 (given)

3y + 3kz = 3 ... equation 2 (multiplied by 3)

3y + (2 + k)z = 1 + k ... equation A

3y + 3kz = 3 ... equation 2 (×3)

——————————<subtract>

(2 + k – 3k)z = k – 2

z = - (k – 2) ⁄ [2(k – 1) ]

... so: k ≠ 1

From this "z" solution and the given equation 2, you can solve for "y" in terms of "k"

Then "x" in terms of "k" can be found using the given equation 1 or 3

The solutions for y, and x also have this: k ≠ 1: restriction

Answer:

There are NO solutions when: k = 1

There is a unique solution for all OTHER real "k" values

Example: for k = 2 , x = -1 , y = 1 , z = 0

so there are an infinite number of unique solutions.

- Login to reply the answers

- lenpol7Lv 77 years ago
y = 1 - kz

Substitute into the other two eq'n.

x + 2(1 - kz) + 2z = 1

-x + 1 - kz + kz = k

x + 2 - 2kz + 2z = 1

-x + 1 = k

Add

2 + 2z (1- k) = 1

1 = k

Substitue

2 + 2z ( k - k ) = 1

2 + 2z(0) = 1

2 + 0 = 1

2 = 1

Answer NO sol'n !!!!

- Login to reply the answers