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October 13th, 2014, 04:29 PM   #1
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Systems of equations

Hello everyone!
Can somebody help me with this? I know its very simple but im beginer and i dont know how to slove it.


A) {2x-y+2z=2
{x+10y-3z=5
{-x+y+z=-3

B) {2x-y+2z-t=2
{x-y+z+t=6
{2x-y-4z+3t=0
{3x+2y-3z+t=2

C) {2x-y+3z=8
{-x+2y+z=4
{3x+y-4z=0
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October 13th, 2014, 04:57 PM   #2
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Quote:
Originally Posted by Cloud View Post
Hello everyone!
Can somebody help me with this? I know its very simple but im beginer and i dont know how to slove it.


A) {2x-y+2z=2
{x+10y-3z=5
{-x+y+z=-3

B) {2x-y+2z-t=2
{x-y+z+t=6
{2x-y-4z+3t=0
{3x+2y-3z+t=2

C) {2x-y+3z=8
{-x+2y+z=4
{3x+y-4z=0
Okay, there are a number of ways to do this problem. But you say you are a beginner so what method(s) have you covered?

-Dan
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October 13th, 2014, 06:00 PM   #3
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Hello, Cloud!

I'll walk through the first one.
I will assume you know the Elimination Method.


Quote:
$(A)\;\begin{Bmatrix}2x-y+2z&=&2 & [1] \\ x+10y-3z&=&5 & [2] \\ \text{-}x+y+z&=&\text{-}3 & [3] \end{Bmatrix}$

$\begin{array}{ccccc}\text{Given [1]:} & 2x - y + 2z &=& 2 \\
2\times [3]\!: & \text{-}2x + 2y + 2z &=& \text{-}6 \\ \hline\end{array}$
$\qquad\text{Add: }\qquad\qquad\; y + 4z \;\;\;=\;\;\;\text{-}4\;\;[4]$

$\begin{array}{cccc}\text{Given [2]:} & x+10y - 3z &=&5 \\
\text{Given [3]:} &\text{-}x\,+\,y\,+\,z &=& \text{-}3 \\ \hline\end{array}$
$\qquad\text{Add: }\qquad\quad 11y - 2z \;\;=\;\;\;\;2\;\;[5]$

$\begin{array}{cccc}\text{Given [4]:} & y + 4z &=& \text{-}4 \\
2\times [5]\!: & 22y - 4z &=& 4 \\ \hline \end{array}$
$\qquad\text{Add: }\qquad\quad\;\; 23y \;\;=\;\;\;0 \quad\Rightarrow\quad y \:=\:0$

Substitute into [5]: $\:11(0) - 2z \:=\:2 \quad\Rightarrow\quad z \:=\:\text{-}1$

Substitute into [2]: $\:x + 10(0) - 3(\text{-}1) \:=\:5 \quad\Rightarrow\quad x \:=\:2$

Therefore: $\:\begin{Bmatrix}x &=& 2 \\ y &=& 0 \\ z&=&\text{-}1 \end{Bmatrix}$
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