Answer:
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
