x+y+z=12 6x-2y+z=16 3x+4y+2z=28 What does x, y, and z equal?

To solve for x, y, and z in the system of equations:

x + y + z = 12 ... (1) 6x - 2y + z = 16 ... (2) 3x + 4y + 2z = 28 ... (3)

We can use the method of substitution or elimination. Here, I will use the elimination method.

First, let's try to eliminate one of the variables from two pairs of equations. We can start by manipulating equations (1) and (2) to eliminate z.

To eliminate z, we can multiply equation (1) by -1 to give us a new equation which we can add to equation (2) to eliminate z:

-1 * (x + y + z = 12) -> -x - y - z = -12 ... (4)

Now, add equations (2) and (4):

(6x - 2y + z = 16) + (-x - y - z = -12) gives us:

6x - 2y + z - x - y - z = 16 - 12 5x - 3y = 4 ... (5)

Now, let's work with equations (1) and (3) to eliminate z again, by multiplying equation (1) by -2 and adding it to equation (3):

-2 * (x + y + z = 12) -> -2x - 2y - 2z = -24 ... (6)

Add equations (3) and (6):

(3x + 4y + 2z = 28) + (-2x - 2y - 2z = -24)

3x + 4y + 2z - 2x - 2y - 2z = 28 - 24 x + 2y = 4 ... (7)

Now we have a simpler system of equations with only x and y:

5x - 3y = 4 ... (5) x + 2y = 4 ... (7)

We can solve equation (7) for x:

x = 4 - 2y ... (8)

Substitute equation (8) into equation (5):

5(4 - 2y) - 3y = 4 20 - 10y - 3y = 4 20 - 13y = 4 -13y = 4 - 20 -13y = -16 y = -16 / -13 y = 16/13

Now plug the value of y back into equation (8) to find x:

x = 4 - 2(16/13) x = 4 - 32/13 x = (52 - 32)/13 x = 20/13

Finally, we can find z using equation (1) with the values of x and y:

(20/13) + (16/13) + z = 12 36/13 + z = 12 z = 12 - 36/13 z = (156 - 36)/13 z = 120/13

So the solution is:

x = 20/13 y = 16/13 z = 120/13