halfway through with substitution problem, need help!

Question

anonymous · Answer

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anonymous · Answer

I used substitution and got y=3x-z+6. then i took that and got the two new equations by plugging it into the other two old equations. I got my two new equations: -3x+4z=26 and 12x-6z=24

anonymous · Answer

I tried to solve for z but ended up getting some weird fraction. not sure what to do next.

anonymous · Answer

@amistre64 can you take a look at this and tell me if i missed a step?

johnweldon1993 · Answer

Hey sorry, working on another question, alright lets see here

$$\large 3x + y + z = 6$$
$$\large 3x - 2y + 2z = 14$$
$$\large 3x + 3y - 3z = -6$$

anonymous · Answer

i can do either elimination or substitution. i started doing substitution but if you want we can do elimination as well

johnweldon1993 · Answer

Ehh, we'll stick with sub since you already started it :)

I believe you solved the first equation for 'y'

so

$$\large 3x + y + z = 6$$
$$\large y = 6 - 3x - z$$

we may have found your mistake early, as I don't see a -3x up there in your post

anonymous · Answer

ohhh i did a positive 3x! okay let me see if that solves my issue.

johnweldon1993 · Answer

Alright, let me know how it goes :)

anonymous · Answer

okay so my new two equations are 9x+4z=26 and -6x-6z=-24

anonymous · Answer

then i can take -6x-6z=-24 and solve for x and i get x=-z+4. so now i will go back and plug that in for x to the other equation.

anonymous · Answer

okay, i know how to finish the problem. thank you!

johnweldon1993 · Answer

Perfect :)