+0

# For a certain value of k, the system

0
115
2
+4

For a certain value of k, the system

x + y + 3z &= 10

-4x + 3y + 5z &= 7

kx + z &= 3

has no solutions. What is this value of k?

Dec 27, 2020

#1
0

The third equation has no y-term, so let's eliminate the y-term from the combination of the first two equations.

x + y + 3z  =  10     --->     x -2     --->     -2x - 2y - 6z  =  -20

-4x + 3y + 5z  =  7       --->                      -4x + 3y + 5z  =  7

Add down the columns:                             -6x - z  =  -13

Since  kx + z  =  3        --->         z  =  3 - kx

Substituting these:        -6x - (2 - 3kx)  =  -15

-6x - 2 + 3kx  =  -15

6x - 2 - 3kx  =  15

6x - 3kx  =  1

3x(2 - k)  =  17

If  2 - k  =  0,  there is no value for x that will result in a product of 10,

so  k = 2  results in no possible solution.

For every other value of k, there will be a value of x that can produce a product of 10.

So the answer is k = 2.

Dec 27, 2020
#2
+117724
+2

Mutiply the first equation through by  -3

-3x  - 3y  - 9z  =  -30       add this to the  second equation

-7x - 4z  = -23

Multiply  the last  equation through by  -4  and we have that

-4kx - 4z  = -12

Note that when k = 7/4   we have the system

-7x -4z =  -23

-7x - 4z = -12

This will have  no solution

So....we will  have no solution when

k  =7/4

Dec 27, 2020