Homework Help, Can you help me by explain what to do to get to the answer, I am stumped because there's only on x variable, so there's not a way to eliminate or substitute, IS THERE?

What does x equal in the solution of the system of equations below?

x + 2y + z = 4
4y – 3z = 1
2y + 10z = 12

x=12.3 or 12 1/3

You take the second and third equations and try to solve them simultaneously.

Using the values of y and z you got there, you use that in the first equation to find the value of x in the first equation.

This is the procedure.

If all three equations have 3 variables, then, you would have to make one variable the subject of formula so that you can get one equation with two variables from two equations (or another way so that you can make two equations with only two variables from 3 equations with 3 variables). This is a little difficult to understand, I know, but if you want an example, I can post one.

Yeah, an example would be great! :]

Ok, let's say:

x + 2y + 3z = 5
2x + y + z = 3
x + 2y + 5z = 9

(just using random numbers, so the results may not be 'pretty numbers'.

From there, you have 3 equations, 3 variables.
Take 1 and 2:
x + 2y + 3z = 5... a
2x + y + z = 3 : x2 -> 4x + 2y + 2z = 6... b

Subtract a from b:
4x - x + 2y - 2y + 2z - 3z = 6-5
3x - z = 1... I

From 2 and 3:
2x + y + z = 3 : x2 --> 4x + 2y + 2z = 6... c
x + 2y + 5z = 9... d

Subtract d from c:
4x - x +2y - 2y +2z- 5z = 6 - 9
3x - 3z = -3... II

From I and II, solve simultaneously, by subtracting II from I:
3x - 3x -3z -(-z) = -3 - 1
-2z = -4
z = 2

Now, use that z in equation I or II to get x.
3x - (2) = 1
x = 1

or

3x - 3z = -3
x - z = -1
x - (2) = -1
x = 1

From the values of x and z, you can find y:
(1) + 2y + 3(2) = 5 : y = -1
2(1) + y + (2) = 3 : y = -1
(1) + 2y + 5(2) = 9 : y = -1

You see? However, surprisingly, I got whole numbers, which isn't always the case when you take random numbers like that.

You could also use matrices. Just involve a 0 variable for the missing link. Ax + by + cz = f

Where x is missing a = 0.