solve the following system of equations?

Question

x - y + 3z = 8, 3x + y - 2z = -2, 2x + 4y + z = 0 (can u guys show me, by breaking it down how u got the answer)

Answer 1

x - y + 3z = 8
3x + y - 2z = -2
2x + 4y + z = 0

The first step is to use any pair of equations to eliminate one of the three variables. I'd choose to add the first two equations vertically, since y and -y will cancel, to yield:

4x + 0y + z = 6
z = 6 - 4x

Now we can substitue "6 - 4x" for "z" in any pair of equations (other than the pair we used to produce the above), to eliminate z and yield two equations for two variables. I'll use the second and third equations (the second equation was used previously, but in combination with the first, so this is a different pair):

3x + y - 2z = -2,
3x + y - 2(6 - 4x) = -2
3x + y - 12 + 8x = -2
11x + y = 10
y = 10 - 11x

2x + 4y + z = 0
2x + 4y + (6 - 4x) = 0
2x + 4y + 6 - 4x = 0
4y = 2x - 6

Now we have two equations and two variables:

y = 10 - 11x
4y = 2x - 6

Substitute the first into the second:

4y = 2x - 6
4(10 - 11x) = 2x - 6
40 - 44x = 2x - 6
46x = 46

x = 1

Now use either of the two equations with just x and y above, and the now known value for x, to solve for y:

4y = 2x - 6
4y = 2*1 - 6
4y = -4
y = -1

Now use any of the original three equations, with the now known values for x and y, to solve for z:

x - y + 3z = 8,
1 - (-1) + 3z = 8
2 + 3z = 8
3z = 6
z = 2

The solution is x=1, y=-1, z=2

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You can check with any or all of the original three equations to verify:

x - y + 3z = 8
1 - (-1) + 3*2 = 8
1 + 1 + 6 = 8
8 = 8

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

Answer 2

x - y + 3z = 8
3x + y - 2z = - 2------ADD

4x + z = 6

4x - 4y + 12z = 32
2x + 4y + z = 0------ADD

6x + 13z = 32

4x + z = 6
6x + 13z = 32

-52x - 13z = - 78
6x + 13z = 32-------ADD

- 46 x = - 46
x = 1

4x + z = 6
4 + z = 6
z = 2

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

x = 1 , y = - 1 , z = 2