Solve the following system of equations using any method 2x+6y+4z=−42 4x+3y+8z=−39 4x+3y+2z=3

the values of x, y and z are: x=8, y = -5 and z = -7

Step-by-step explanation:

2x+6y+4z=−42       eq(1)

4x+3y+8z=−39       eq(2)

4x+3y+2z=3          eq(3)

We would solve the above equations using elimination method.

Subtracting eq(3) from eq(2)

4x+3y+8z=−39      

4x+3y+2z=3

-   -     -       -

0+0+6z = -42

z = -42/6

z = -7

Multiplying eq(1) with 2 and subtracting with eq(2)

4x + 12y +8z = -84

4x +3y +8z = -39

-    -      -        +

0+9y+0=-45

9y = -45

y = -45/9

y = -5

Putting value of y and z in eq(1)

2x + 6y +4z = -42

2x + 6(-5) +4(-7) = -42

2x -30 -28 = -42

2x -58 = -42

2x = -42 +58

2x = 16

x = 16/2

x= 8

So, the values of x, y and z are: x=8, y = -5 and z = -7