What is the solution to the following system? 4x+3y-z=-6 6x-y+3z=12 8x+2y+4z=6 a. x = 1, y = –3, z = –1 b. x = 1, y = –3, z = 1 c. x = 1, y = 3, z = 19 d. x = 1, y = 3, z = –2

Option B is correct.

Step-by-step explanation:

4x+3y-z= -6     eq(1)

6x-y+3z= 12     eq(2)

8x+2y+4z=6    eq(3)

We need to solve these equations and find the value of x, y and z.

Multiply equation 2 with 3 and then add equation 1 and 2

18x -3y +9z = 36

4x +3y -z = -6

22x + 8z = 30    eq(3)

Multiply equation 2 with 2 and add with equation 3

12x -2y + 6z = 24

8x +2y +4z = 6

20x + 10 z = 30  eq(4)

Multiply equation 3 with 10 and equation 4 with 8 and then subtract

220x + 80z = 300

160x + 80z = 240

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60x = 60

x= 60/60

x= 1

Putting value of x in equation 3

22x + 8z = 30

22(1) + 8z = 30

8z = 30 - 22

8z = 8

z = 8/8

z=1

Putting value of x and z in equation 1

4(1)+3y-(1)=-6

4 + 3y -1 = -6

3 + 3y = -6

3y = -6 -3

3y = -9

y = -3

so, Option B x=1, y=3 and z=1 is correct

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