Question

Solve the following system of equations:

2x + 3y − z = 1
3x + y + 2z = 12
x + 2y − 3z = −5

Answer 1

Which method are you using in class? Are you doing matrices? Or elimination of a variable?

Answer 2

I can use elimination or substitution.

Answer 3

Ok, well elimination is a bunch easier! Can we do that?

Answer 4

yes

Answer 5

The goal is to take the first 2 equations and eliminate a variable, then take the second and third equation and eliminate the same variable, then the two resulting equations, each of them minus one variable can be solved simultaneously for x or y or z. Then use that to find the other variables. Let's start with eliminating the z. Cuz z's are weird in equation, ok? ; )

Answer 6

We will eliminate the z between the first two equations:
2x + 3y - z = 1
3x + y + 2z = 12

Answer 7

We have to make the coefficients the same number, but one of them negative and the other positive so they cancel each other out. In order to do that, let's make both the coefficients in front of those z's a 2.

Answer 8

To do that, multiply the first equation by a 2 like this:

Answer 9

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

Answer 10

Now let's add that to the second equation and see what happens:

Answer 11

4x + 6y - 2z = 2
3x + y + 2z = 12
----------------
7x + 7y + 0z = 14
Of course 0z is equal to no z's at all, so what we are left with is:
7x + 7y = 14

Answer 12

That's the first 2 equations. Now let's do equation 2 and 3 the same way:

Answer 13

3x + y + 2z = 12
x + 2y - 3z = -5

Answer 14

In order to eliminate the z's in these two, we have to again get the coefficients the same, but one negative and one positive. Between a 2 and a 3, 6 is the LCM, so we have to multiply the second equation by a 3, and the third equation by a 2 to make them both 6's. Like this:

Answer 15

3(3x + y + 2z = 12)
2( x + 2y - 3z = -5)

Answer 16

Multiplying those through you get this:

Answer 17

9x + 3y + 6z = 36
2x + 4y - 6z = -10
-----------------
11x + 7y + 0z = 26
Of course, again, the 0z is gone, so we have two new equations now, which are:

Answer 18

7x + 7y = 14
11x + 7y = 26

Answer 19

Now we have to work to eliminate another variable, which is easy this time...there are 7's in front of the both y coordinates, but both of them are positive. We have to make one of them negative, so we will multiply either one by a -1, like this (I am going to pick the top one to multiply by a -1):

Answer 20

-1(7x + 7y = 14) --> -7x - 7y = -14

Answer 21

Adding that to the bottom equation we now have this:

Answer 22

-7x - 7y = -14
11x + 7y = 26
-------------
4x + 0y = 12 or
4x = 12.

Answer 23

Solve that for x and get that x = 3. NOW fill the 3 back in to one of our equations that has only an x and a y in it to get this (I m going to use the equation 7x + 7y = 14):

Answer 24

7(3) + 7y = 14 --> 21 + 7y = 14 --> 7y = 14 - 21 --> 7y = -7 --> y = -1.

Answer 25

Now we have x = 3 and y = -1. Fill those into one our original equations with all three variables to solve for z:

Answer 26

2x + 3y - z = 1 --> 2(3) + 3(-1) - z = 1 --> 6 - 3 - z = 1 --> 3 - z = 1 --> -z = 1 - 3 --> -z = -2 or z = 2.

Answer 27

Now you have all your variables: x = 3, y = -1, z = 2. See that?

Answer 28

yes! thank you!

Answer 29

You are welcome!