October 13th, 2014, 04:29 PM  #1 
Newbie Joined: Oct 2014 From: macedonia Posts: 4 Thanks: 0  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) {2xy+2z=2 {x+10y3z=5 {x+y+z=3 B) {2xy+2zt=2 {xy+z+t=6 {2xy4z+3t=0 {3x+2y3z+t=2 C) {2xy+3z=8 {x+2y+z=4 {3x+y4z=0 
October 13th, 2014, 04:57 PM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 928 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
October 13th, 2014, 06:00 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, Cloud! I'll walk through the first one. I will assume you know the Elimination Method. Quote:
$\begin{array}{ccccc}\text{Given [1]:} & 2x  y + 2z &=& 2 \\ 2\times [3]\!: & \text{}2x + 2y + 2z &=& \text{}6 \\ \hline\end{array}$ $\qquad\text{Add: }\qquad\qquad\; y + 4z \;\;\;=\;\;\;\text{}4\;\;[4]$ $\begin{array}{cccc}\text{Given [2]:} & x+10y  3z &=&5 \\ \text{Given [3]:} &\text{}x\,+\,y\,+\,z &=& \text{}3 \\ \hline\end{array}$ $\qquad\text{Add: }\qquad\quad 11y  2z \;\;=\;\;\;\;2\;\;[5]$ $\begin{array}{cccc}\text{Given [4]:} & y + 4z &=& \text{}4 \\ 2\times [5]\!: & 22y  4z &=& 4 \\ \hline \end{array}$ $\qquad\text{Add: }\qquad\quad\;\; 23y \;\;=\;\;\;0 \quad\Rightarrow\quad y \:=\:0$ 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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