Answer:
values of x,y and z are x = -2, y= -9 and z=5
Step-by-step explanation:
2x+4y+1z=−35 eq(1)
3x+7y+7z=−34 eq(2)
2x+10y+6z=−64 eq(3)
We can solve using elimination method
Subtracting eq (1) from eq(3)
2x + 10y +6z = -64
2x +4y +1z = -35
______________
6y + 5z = -29 eq(3)
Multiplying eq(2) with 2 and eq(3) with 3 and subtracting
6x + 14y +14z = -68
6x + 30y +18z = -192
- - - +
_________________
-16y -4z = 124 eq(4)
Multiply eq(3) with 4 and eq(4) with 5 and add both equations
24y + 20z = -116
-80y - 20z = 620
______________
-56y = 504
y = -504/56
y= -9
Putting value of y in equation(3)
6y + 5z = -29
6(-9) + 5z = -29
-54 + 5z = -29
5z = -29+54
5z = 25
z = 25/5
z =5
Now, putting value of y and z in eq(1)
2x + 4y +1z = -35
2x + 4(-9) +1(5) = -35
2x -36+5 = -35
2x -31 = -35
2x = -35+31
2x = -4
x= -4/2
x=-2
So, values of x,y and z are x = -2, y= -9 and z=5
