Solve the following system of equations: 2x + 4y − 3z = −7 3x + y + 4z = −12 x + 3y + 4z = 4

2x + 4y − 3z = −7 ---> first equation

3x + y + 4z = −12   ---> second equation

x + 3y + 4z = 4   ---> third equation

We try to eliminate z  and make two equations

Multiply first equation by 4 and second equation by 3. then add both equations

8x + 16y -12z = -28

9x + 3y +12z = -36

17x + 19y = -64   > equation 4

multiply third equation by -1  and add second and third equation

3x + y + 4z = −12  

-x - 3y - 4z = -4    

2x -2y = -16        > equation 5

Multiply equation 4 by 2 and equation by 19

34x + 38y = -128

38x - 38y = -304

72x  = -432

Divide by 72 on both sides

x = -6

use equation 2x -2y = -16        > equation 5

2(-6) - 2y = -16

-12 -2y = -16

-2y = -4

So y = 2

Now plug in 2 for y  and -6 for x

Using any one of the three equations

x + 3y + 4z = 4  

-6 + 3(2) +4z = 4

4z = 4 (divide by 4 on both sides)

So z=1

So x= -6, y = 2  and z= 1

whatvare the choices

step-by-step explanation: