Determine which of () form a solution to the given system for any choice of the free parameter. 3x_1 + 8x_2 - 14x_3 = 9 x_1 + 3x_2 - 4x_3 = 1 a. (6 - 2s_1, 5 + 3s_1, s_1) a. solutionb. not a solutionb. (-11 - 5s_1, s_1 - (6 +s_1)/2)a. solution
b. not a solution c. (19 + 10s_1, -6 - 2s_1, s_1)a. solution
b. not a solution d. ((5 -4s_1)/3, - s_1)/4)a. solution
b. not a solution

a = Not a solution

b = Solution

c = Solution

d = Not a solution

Step-by-step explanation:

Here we are given with two equations so lets call them equation 1 and 2

3x_1 + 8x_2 - 14x_3=9(1)

 x_1 + 3x_2 - 4x_3 = 1(2)

Part (a)

For values: (6-2s_1, 5+3s_1 , s_1)

Putting the values in Equation (2) for convenience however it can also be solved by substituting in Equation (1).

6-2s_1 + 3(5+3s_1) - 4s_1  = 1

6-2s_1 + 15 + 9s_1 - 4s_1 = 1

21+3s_1 = 1

L.H.S ≠ R.H.S

Hence not a solution

Part (b)

For values: (-11-5s_(1,) s_1,(-(6+s_1 ))/2)

11-5s_1 + 3s_1 - 4(-(6+s_1)/2) = 1

-11-5s_1 + 3s_1 + 12+2s_1  = 1

1 = 1

L.H.S = R.H.S

Hence it has a solution

Part (c)

For values:(19+10s_1,-6-2s_1,s_1)

19 + 10s_1 + 3(-6-2s_1) - 4s_1 = 1

19 + 10s_1 - 18 - 6s_1  - 4s_1 = 1

1 = 1

L.H.S = R.H.S

Hence it has a solution

Part (d)

For values:  ((5-(4s_1)/3 , s_1, -(7-s_1)/4)

((5-4s_1)/3) + 3s_1  - 4(-(7-s_1)/4))  = 1

5/3 - 4/3s_1 + 3s_1 + 7-s_1  = 1

5/3 + 7- (4/3s_1) + 2s_1 = 1

L.H.S ≠ R.H.S

There is no need to simplify this any further as it can be seen it has no solution.

the reason why your question keeps getting deleted because this is not a school related question : )

i will answer your question though.

i never tried church's

answer: 20

step-by-step explanation: