system of equations?

Question

solve:

-2x-2y+4z=12
3x-y-3z=-2
2x-y-2z=-1

I know the method you are supposed to use, but when i check my answer it doesnt seem to come out right. If you could show me how u come up with the answer it would really help me to understand it, thanks.

thanks alot, i totally get it now!!!

Answer 1

-2x - 2y + 4z = 12
3x - y - 3z = -2
2x - y - 2z = -1

Take the first two equations:
-2x - 2y + 4z = 12
3x - y - 3z = -2

multiply the 2nd one by -2
-2x - 2y + 4z = 12
-6x + 2y + 6z = 4
----------------------
-8x +10z = 16

Take the last two equations:
3x - y - 3z = -2
2x - y - 2z = -1

multiply the 2nd one by -1
3x - y - 3z = -2
-2x + y + 2z = 1
---------------
x - z = -1

Take this with the equation from above and do system of equations:
-8x +10z = 16
x - z = -1

Multiply by 8
-8x +10z = 16
8x - 8z = -8
-------------------
2z = 8
z = 4


x - z = -1
x - 4 = -1
x = 3


-2x - 2y + 4z = 12
-2(3) - 2y + 4(4) = 12
-6 - 2y + 16 = 12
-2y + 10 = 12
-2y = 2
y = -1

-------------

x = 3
y = -1
z = 4

Hopefully you can understand this :)

Answer 2

1) - 2x - 2y + 4z = 12
2) 3x - y - 3z = - 2
3) 2x - y - 2z = - 1

You want to produce two equations in two unknowns - there are various ways you could do this

note that you can divide both sides of equation 1 by 2 with no harm

1a) - x - y + 2z = 6
2) 3x - y - 3z = - 2
3) 2x - y - 2z = - 1

Since the coefficient of y is now 1 for all equations, subtract equation 2 from equation 1, and subtract equation 3 from equation 2

4) - 4x + 5z = 10
5) x - z = - 1

Solve equation 5 for x and substitute into equation 4

5) x - z = - 1
x = z - 1

4) - 4x + 5z = 8
- 4(z - 1) + 5z = 8
- 4z + 4 + 5z = 8
z = 4

Use equation 5 to get x

5) x - z = - 1
x - 4 = - 1
x = 3

Plug the x & z values into any original equation to get y

1) - 2x - 2y + 4z = 12
- 2(3) - 2y + 4(4) = 12
- 6 - 2y + 16 = 12
- 2y = 2
y = - 1

Time permitting, check using the other original equations

2) 3x - y - 3z = - 2
3(3) - (-1) - 3(4) = - 2
9 + 1 - 12 = - 2
- 2 = - 2

etc.