Question

PLEASE help me practice solving systems 3 equations, 3 variables, 3 answers. PLEASE help me practice solving systems?? I get to re-take a systems quiz tomorrow and I want to do better on it. PLEASE HELP will REWARD AN MEDAL..

Answer 1

I don't have a particular problem. I'm wanting help solving sysems? with 3 equations... I don't have a particular problem?

Answer 2

http://www.youtube.com/watch?v=aPKz7011lAA maybe ?

Answer 3

doesn't help me

Answer 4

hmm... is pretty thorough IMO

Answer 5

lemme cook one up then

Answer 6

in general use elimination to eliminate a variable and reduce it to a system of 2 equations , then solve that system , then get 3rd variable by back substituting

Answer 7

maybe this video will help
https://www.khanacademy.org/math/algebra/systems-of-eq-and-ineq/fancier-systems/v/systems-of-three-variables

Answer 8

\(\large \begin{array}{llll}
2x-2z+3y=-24\\
5z+3y=27\\
5y-3z=-47
\end{array}\)

Answer 9

hmm I have a typo

Answer 10

\(\large \begin{array}{llll}
2x-2z+3y=-24\\
5z+3x=27\\
5y-3z=-47
\end{array}\)

Answer 11

so first off... we want to order all variables, just like in a 2 variable set, in a column

notice the 2nd and 3rd rows, they seem to be missing a variable, thus we PAD it with 0
since 0x = " "

Answer 12

so we end up with then \(\large \begin{array}{llll}
2x+3y-2z=-24\\
3x+0y+5z=27\\
0x+5y-3z=-47
\end{array}\)

Answer 13

so as the video showed, we start by grabbing a pair first from the 3 available
then we cancel out a variable, say "z"

Answer 14

2x+3y-2z=-24 <---- let us pick this one
3x+0y+5z=27 <--- and this one for now
0x+5y-3z=-47

Answer 15

\(\large {\begin{array}{llll}
2x+3y-2z=-24 & \times 5\implies &10x+15y-10z=-120\\
3x+0y+5z=27& \times 2\implies &6x+0y+10z=54\\
\hline\\
&&16x+15y+0z=-66
\end{array}\\ \quad \\
\color{blue}{16x+15y=-66}}
\)

Answer 16

so by using the elimination method, we ended up with the one above, our 1st 2 variable function

Answer 17

@jdoe0001 I understand it all up until ur entire last message I don't understand ur last message sir/mam????

Answer 18

ok... .so... what part confused you?

Answer 19

the right side how u transformed the problem to the new two problems??

Answer 20

\(\large \begin{array}{llll}
2x+3y-2z=-24 & \huge{\color{red}{\times 5}}\implies &10x+15y-10z=-120\\
3x+0y+5z=27& \huge {\color{red}{\times 2}}\implies &6x+0y+10z=54\\
\hline\\
&&16x+15y+0z=-66
\end{array}\\ \quad \\
16x+15y=-66
\)

Answer 21

see it now?

Answer 22

the idea being, I'd end up with a "z" value atop, EQUAL BUT DIFFERENT SIGN as the one at the bottom

Answer 23

yupp now I understand I undertand it now thanks

Answer 24

so let us pick another 2 functions from the available 3

2x+3y-2z=-24 <--- this one again
3x+0y+5z=27
0x+5y-3z=-47 <---- and this one this time

Answer 25

I undertand it now I watched the videos thanks so much for ur help... and u explaining it helped too. I think get it now

Answer 26

\(\large {\begin{array}{llll}
2x+3y-2z=-24 & \color{red}{\times -3}\implies &-6x-9y+6z=72\\
0x+5y-3z=-47& \color{red}{\times 2}\implies &0x+10y-6z=-94\\
\hline\\
&&-6x+1y+0z=-22
\end{array}\\ \quad \\
\color{blue}{-6x+y=-22}}\)

Answer 27

so there, then you use the 2 resulting 2 variables functions and solve for either
then get the other and then plug them back in any of the originals and to get "z" :)

Answer 28

(-= THANKS SO MUCH..

Answer 29

yw