Question

Using complete sentences, explain which method you would use to solve the following system of equations and why. In your answer, include the solution to one of the variables and how you found it using the method you chose.

x – 5y + 2z = 0
x + 4y – z = 12
2x – y + 3z = 10

Answer 1

I would set up the 3x4 matrix and reduce it to reduced row echelon form to solve the system of equation.

Now that I've answered that question for you, do you have any questions about that question?

Answer 2

Do you have time to do this step by step?

Answer 3

The method I used is matrix algebra. Have you learned how to do that yet?

Answer 4

Isn't that where you have to use two eliminate two equations solve for the variable then use the answer that i get from that to plug it into the equation that i didn't use in the beginning?

Answer 5

\[\left[\begin{matrix}1 & -5 & 2 & 0 \\ 1 & 4 & -1 & 12\\ 2 & -1 & 3 & 10\end{matrix}\right]\] is the matrix that you want to set up.
Reduced row echelon form is \[\left[\begin{matrix}1 & 0 & 0 & a \\ 0 & 1 & 0 & b\\ 0 & 0 & 1 & c\end{matrix}\right]\]
where \[\left[\begin{matrix}x & y & z & constant \end{matrix}\right]\]
so the solution is x =a , y =b z= c

Answer 6

so they all = 1?

Answer 7

you reduce it to 1.
You take the second row, and subtract it from the first (1 4 -1 12) - (1 -5 2 0) and you get (0 9 -3 12)
You take the third row, and subtract it from two times the first row (2 -1 3 10) - 2*(1 -5 2 0) and you get 0 9 -1 0)
and now you're left with this matrix
\[\left[\begin{matrix}1 & -5 & 2 & 0 \\ 0 & 9 & -3 & 12\\ 0 & 9 & -1 & 10\end{matrix}\right]\]
the subtract the third row from the second row
\[\left[\begin{matrix}1 & -5 & 2 & 0 \\ 0 & 9 & -3 & 12\\ 0 & 0 & 2 & -2\end{matrix}\right]\]
Divide the third row by 2
\[\left[\begin{matrix}1 & -5 & 2 & 0 \\ 0 & 9 & -3 & 12\\ 0 & 0 & 1 & -1\end{matrix}\right]\]
and now you have a simpler system of equations
x - 5y +2z = 0
0x+9y -3z = 12
0x + 0y +z = -1
Then substitute the rest or continue manipulating the matrix

Answer 8

ooh okay thanks :)

Answer 9

Have you had that method yet @Yoduh ?

Answer 10

To be honest i haven't but thing is i want to do this using the Elimination method.
Think you can help me?

Answer 11

The elimination method is certaInly a method that could be used.

x – 5y + 2z = 0
x + 4y – z = 12
2x – y + 3z = 10

Answer 12

Note that the first two equations can be used to eliminate x.
as you can do this x=x or
5y-2z=-4y + z + 12 or
9y - 3z - 12 = 0 see the x is gone now we need to eliminate y or z. Did you see how the x was eliminated?

Answer 13

yes i know how to do that part just it gets a little confusing when i want to eliminate whichever other variable. I get two variables on one side and a regular number on the other

Answer 14

Give me one second I'm going to do this in equation form so you can see if I'm doing it right

Answer 15

x-5y+2z=0 and x+4y-z=12

I need to eliminate the x so want to multiply which ever equation by -1

-1(x-5y+2z=0) making it -x+5y-2z=0

then i add them

-x+5y-2z=0
+ x+4y-z=12
9y-z=12
am I correct?

Answer 16

or is t 9y-3z=12

Answer 17

the latter

Answer 18

addition of (-2z and -z)= -3z

Answer 19

You have now eliminated x and have the equation 9y-3z=12

Answer 20

great but this is what i mean two variables on one side? how do i fix that into making it to one variable so i can use it as part of the equation?

Answer 21

As a suggestion take the first and last equation and eliminate x again giving you another new equation with only y and z terms.

Answer 22

i got this
X+5y+2z=0
+2x-y+3z=10
= 3x-4y+5z=10