Systems of Linear Equations in 3 variables

[Ms. Gidget]

I am getting pretty mixed up trying to solve this one. Could someone show me the steps to figuring out this Linear Equation with 3 variables?

-5x + 3y +2z = 1
x + y + z = 7
2x - y + z = 7

 

[tkhunny]

There are several methods. The most important element of ANY method is paying attention. These can get frustrating and confusing. The neater and cleaner your mothodology, the more likely you are to succeed.

What methods have you studied? Substitution usually is first.

1) -5x + 3y +2z = 1
2) x + y + z = 7
3) 2x - y + z = 7

Solve ANY of the equations for ANY of the variables. Try to pick an easy one. it will get complicated soon enough, so no need to make it worse than it is. I'll choose 2) and solve for x.

4) from 2) x = 7 - y - z <== We'll need this later. Keep track of it.

Substitute this value of 'x' into the other 2 equations.

5) Substituting into 1) -5(7 - y - z) + 3y + 2z = 1
6) Substituting into 3) 2(7 - y - z) - y + z = 7

Simplfy things and start over.

7) Simplifying 5) 8y + 7z = 36
8) Simplifying 6) 3y + z = 7

9) from 8) z = 7 - 3y <== We'll need this later. Keep track of it.

10) Substituting into 7) 8y + 7(7 - 3y) = 36

11) Simplifying 10) y = 1
12) Substituting into 9) z = 7 - 3(1) = 7 - 3 = 4
13) Substituting into 4) x = 7 - 1 - 4 = 7 - 5 = 2

14) Checking Solution in 1) -5(2) + 3(1) + 2(4) = -10 + 3 + 8 = -7 + 8 = 1 as required.
15) Checking Solution in 2) (2) + (1) + (4) = 2 + 1 + 4 = 3 + 4 = 7 as required.
16) Checking Solution in 3) 2(2) - (1) + (4) = 4 - 1 + 4 = 3 + 4 = 7 as required.

I think we're done.

Note: I skipped a few steps. Truthfully, is put the additional steps on a separate piece of paper. Don't try to do it in your head. Let the notation help you. If you can't keep track with your present level of detail, use more detail. Don;t be afraid to write things down and recap your progress at each state.

note: Extra Stress on WRITING DOWN EVERYTHING.

 

[Mrspi]

I am getting pretty mixed up trying to solve this one. Could someone show me the steps to figuring out this Linear Equation with 3 variables?

-5x + 3y +2z = 1
x + y + z = 7
2x - y + z = 7

Here's another approach....

If you can turn this system of three equations in three variables into a system of two equations in two variables, that should simplify things...

Add equations 2 and 3 from the given system:

2).... x + y + z = 7
3).... 2x - y + z = 7
---------------------------
....... 3x + 2z = 14

Ok...now you have one equation that contains just x and z as variables. Can you get another one like this??

I believe you can. Multiply the third equation by 3....
3(2x - y + z) = 3*7
6x - 3y + 3z = 21

ADD this equation to equation 1:

-5x + 3y + 2z = 1
6x - 3y + 3z = 21
-----------------------
x + 5z = 22

Ok....now you have TWO equations in two variables:

3x + 2z = 14
x + 5z = 22

You should be able to solve a system of two equations with two variables. Try it.

If you're still having trouble with this problem, please repost, showing us all of the work you've done. When we can see your work, we can better determine how to help you.

 

[galactus]

Here is a method known as RREF (reduced row echelon form). The idea is elimination.

Start with your system only write it without the variables:

\(\displaystyle \begin{bmatrix}-5&3&2&1\\1&1&1&7\\2&-1&1&7\end{bmatrix}\)

The idea is to get it into the form \(\displaystyle \begin{bmatrix}1&0&0&a\\0&1&0&b\\0&0&1&c\end{bmatrix}\)

Where a, b, and c are the solutions to your system>

Now, let's multiply row 2 by -2 and add it to row 3, then we get:

\(\displaystyle \begin{bmatrix}-5&3&2&1\\1&1&1&7\\0&-3&-1&-7\end{bmatrix}\)

Now, multiply row 1 by 1/5 and add to row 2, then we get:

\(\displaystyle \begin{bmatrix}-5&3&2&1\\0&8/5&7/5&36/5\\0&-3&-1&-7\end{bmatrix}\)

Now, multiply row 2 by 5/8, then we get:

\(\displaystyle \begin{bmatrix}-5&3&2&1\\0&1&7/8&9/2\\0&-3&-1&-7\end{bmatrix}\)

Now, multiply row 2 by -3 and add to row 3, then we get :

\(\displaystyle \begin{bmatrix}-5&3&2&1\\0&1&7/8&9/2\\0&0&13/8&13/2\end{bmatrix}\)

Now, multiply row 3 by 8/13:

\(\displaystyle \begin{bmatrix}-5&3&2&1\\0&1&7/8&9/2\\0&0&1&4\end{bmatrix}\)....See there?. we have z=4 in the last row.

Now, multiply row 3 by -7/8 and add to row 2, then we get:

\(\displaystyle \begin{bmatrix}-5&3&2&1\\0&1&0&1\\0&0&1&4\end{bmatrix}\)

Multiply row 2 by -3 and add to row 1, then we get:

\(\displaystyle \begin{bmatrix}-5&0&2&-2\\0&1&0&1\\0&0&1&4\end{bmatrix}\)

Multiply row 1 by -1/5, then we get:

\(\displaystyle \begin{bmatrix}1&0&-2/5&2/5\\0&1&0&1\\0&0&1&4\end{bmatrix}\)

Now, the last one finally. multiply row 3 by 2/5 and add to row 1, then we get:

\(\displaystyle \begin{bmatrix}1&0&0&2\\0&1&0&1\\0&0&1&4\end{bmatrix}\)

So, the solutions are \(\displaystyle \fbox{x=2, \;\ y=1, \;\ z=4}\)

There is no set way to do this. It is just a matter of whittling away at it and trying not to replace something you already done.

I know it seems tedious and long. That's because it is.

This method takes practice , like everything else.