Question

Need some help with solving systems of linear equations in three variables?
3x+y+z=14
-x+2y-3z=-9
5x-y+5z=30

Answer 1

This video shows how to do elimination
http://www.khanacademy.org/math/linear-algebra/v/solving-3-equations-with-3-unknowns

Answer 2

yeah i watched it and dont get it at all. I would like someone mostly just to solve the equation i posted.

Answer 3

You can do that two ways, one is the old fashioned algebra 1 way where you eliminate variables (in this case, hold two constant) to solve for the third, and then substitute. The second is to use a matrix and do the Gaussian elimination using the Gauss-Jordan algorithm.

Answer 4

I'm guessing they want you to use Gaussian elimination? If so, start by writing the coefficients in an augmented matrix:

\[ \huge
\begin{bmatrix}
3 & 1 & 1 & 14 \\
-1 & 2 &-3 & -9 \\
5 & -1 & 5 & 30
\end{bmatrix}
\]

You can see all of the relevant steps here: http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi by clicking "Transforming a matrix to row echelon form".

Answer 5

3x+y+z=14
3(-x+2y-3z)=-9*3
add both eqation get 7y -8z=-13
3( 5x-y+5z)=3*30
5(3x+y+z)=5*14
subtract both eqa we get -8y+10z=20
again 8(7y-8z)=-13*8
7(-8y+10z)=7*20
add both of them 6z=36
z=6( put in -8y+10z=20 get y=5)
y=5
x=1