Question

Solve the following linear expressions by

elimination:

2x+y+z=1
x-y+2z=5
3x+4y-z=0

Answer 1

may you help me out wtih this?

Answer 2

have you done elimination before?

Answer 3

yes

Answer 4

I get 21x+21z=42
-21x-21z=-60

0 does not equal -18

Answer 5

so is the solution trivial?

Answer 6

ok.... so..... lemme show you how to do it for 3 variables
so we take 2 functions at a time
eliminate one of the 3 variables
then we end up with a 2 variable set
then we do the same thing again
then we end up with 2 variables system of equations


2x+y+z=1 <---- let's use this one
x-y+2z=5 <---- and this one this time
3x+4y-z=0

Answer 7

gimme a sec

Answer 8

ty, jdoe, you are the best

Answer 9

you still there?

Answer 10

yes hehee, I was typing..... ...just not here :)

Answer 11

ty

Answer 12

\(\begin{array}{llrll}
2x+y+z=1& \times -2\implies &-4x-2y-2z=-2\\
x-y+2z=5&&x\quad -y\quad +2z=5\\
\hline\\
&&-3x\quad -3y\quad +0z=3\\ \quad \\
\bf \color{blue}{-3x-3y=3}
\end{array}\)

Answer 13

hmm

Answer 14

so that's our first 2 variable function

now for our next one

2x+y+z=1 <-- let's use this one again
x-y+2z=5
3x+4y-z=0 <--- and this one this time

one sec

Answer 15

same problem, it cancels out to 0 does not equal 18 :/

Answer 16

\(\begin{array}{rllrll}
2x+y\quad +z=1&\textit{notice we don't have to multiply by anything}\\
3x+4y-z=0\\
\hline\\
5x+5y+0z=1\\ \quad \\
\bf \color{blue}{5x+5y=1}
\end{array}\)

Answer 17

yeah

Answer 18

but

Answer 19

then you have to add up -3x-3y=3 +5x+5y=1

Answer 20

and they cancel out to 0 does not equal 18

Answer 21

hmm

Answer 22

lemme check that

Answer 23

ty, jdoe, you are a HERO

Answer 24

hmm you're right.... it ends up as 0 = 18
that means that the system is inconsisten, and thus it has no solution, meaning the 3 lines do not touch each other

Answer 25

jdoe

Answer 26

if it has one solution only, is it Independent and consistent?

Answer 27

yes

Answer 28

ty

Answer 29

yw