Question

x+y+z=2 y-3z=6 2x-3y+z=6 HELP!!

Answer 1

there is no easy way to do this other than cheating

Answer 2

Can I get help step by step?

Answer 3

Get all the x values together,all y together and all numericals together and all z together.

Answer 4

we can try
i guess you have to use elimination

Answer 5

Also if you're shifting after = <-- sign changes

Answer 6

alright, I'm not sure how exactly to solve this.. I'll appreciate it!

Answer 7

If I get help step by step. (:

Answer 8

\[x+y+z=2\\ y-3z=6 \\2x-3y+z=6\]

Answer 9

Is that Cramers rule? @satellite73

Answer 10

lets get rid of the \(x\) first by taking \[x+y+z=2\\2x-3y+z=6\] and multiply the first equation by \(-2\) to get
\[-2x-2y-2z=-4\\2x-3y+z=6\]

Answer 11

You can use elimination to get rid of x in the first and third equations. Then you have a simpler system of two equations with y and z only. But there's many ways to do this.

Answer 12

add \[-2x-2y-2z=-4\\2x-3y+z=6\] and get
\[-5y-z=2\]now the \(x\) is gone

Answer 13

now looking at the second equation and the one above, we get two equations in two unknowns
\[-5y-z=2\\y-3z=6\]

Answer 14

solving this almost in your head you get \(y=0, z=-2\) but we can work out those steps too

Answer 15

alrighty (:

Answer 16

\[-5y-z=2\\y-3z=6\] multiply the second one by 5 get \[-5y-z=2\\5y-15z=30\]
add and get \(-16z=32\iff z=-2\)

Answer 17

then substitute back to find \(y-3(-2)=6\implies y+6=6\implies y=0\) as promised

Answer 18

now you have \(y=0, z=-2\) substitute in to
\[x+y+z=2\] and get
\[x+0-2=2\implies x=4\]

Answer 19

solution is \((4,0,-2)\) and i hope i never have to do another one of these as long as i live

Answer 20

haha thank you so much!

Answer 21

if it was me, i would do this
http://www.wolframalpha.com/input/?i=x%2By%2Bz%3D2%2C+y-3z%3D6%2C+2x-3y%2Bz%3D6+
done is 2 seconds, and have a snack

Answer 22

yw

Answer 23

thank you again!! i have more to do but i kinda got the idea!

Answer 24

@satellite73 How do I use that website?