Question

WILL GIVE MEDAL FOR BEST ANSWER!
Solve the system by elimination.

{-2x+2y+3z=0
{-2x-y+z=-3
{2x+3y+3z=5

Answer 1

Super simple stuff. Just solve two, then solve the last one by plugging in your x-y values

Answer 2

How do I start? I have a general idea of how to do this, I'm just confused about a couple of things.

Answer 3

By the way, imagine that they are all in one big bracket. That's how it's supposed to be, but I don't know how to do that on here.

Answer 4

hmmm have you done system of equations yet? with 2 variables? how abotu with 3 variables?

Answer 5

I have done a system of equations before, with 3 variables and 2 variables. I'm a little rusty though, and don't really remember how to start...

Answer 6

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

We need to eliminate our first equation. We can do this by eliminating z.

Multiply out second equation by -3.

-2x + 2y + 3z = 0
6x + 3y - 3z = 9
2x + 3y + 3z = 5

We we can eliminate z by adding

4x + 5y = 9
2x + 3y + 3z = 5

So now we do the same thing for 2x + 3y + 3z = 5. Eliminate it and remove z, then you'll hve a system with two (,x, y) points you can solve.

Answer 7

well... pick any two equations from the three available ones
and cancel out 1 of the variables

Answer 8

then you pick another set of 2 off the three available
and cancel out the SAME variable

then use the two RESULTANT equations from each cancellation
and solve it, like you would any system of equations of two variables

Answer 9

Sooo I think the way you're supposed to do it is to add/subtract 2 equations from each other, and then use the result of those 2 equations to solve for one or 2 variables, right? Is that the way you guys are doing it?

Answer 10

After that, plug in both your x and y values to find z. Tada!

Answer 11

So you cancel out z first, right? I'm a little confused about the way you wrote it...

Answer 12

You multiplied the second equation by -3 and then subtracted that equation from both of the other equations to cancel out the z?

Answer 13

(1)+(3).......(4)
(2)+(3).......(5)

(4) and (5) contain only y and z, i.e. 2x2.

Answer 14

@mathmate Ohh the way you wrote it is much easier to understand, no offense. Okay, I'll try it your way.

Answer 15

It is quite true that eliminating any variable will work. Sometimes you will notice that certain variables are easier to be eliminated than others! :)

Answer 16

@mathlete
Alright, so equations (4) and (5) are:
{5y+6z=5
{2y+4z=2

Sooo... now what? Now do you just solve them like a two variable equation? I kinda forgot how to do this a little...

Answer 17

@mathmate
I got z=0. Is that wrong? It seems wrong...

Answer 18

z=0 is correct, but you still have to find the others.

Answer 19

Okay, good! I was just wondering if I did it right. So how do I find the others now?

Answer 20

Choose any two of the equation (say the first two), put z=0 and solve again a system of two unknowns (x and y).

Answer 21

@mathmate
What 2 equations could I use? The original equations, or equations (4) and (5)?

Answer 22

Equation (4) and (5) do not contain x any more, so you must use two of (1), (2) or (3).

Answer 23

Okay, but could I plug (z=0) into (4) or (5) to get the value for y?

Answer 24

@mathmate

Answer 25

Yes, that's even better, lol!

Answer 26

Yay! I think I can figure out the rest then. Thank you so much for all your help! :)

Answer 27

Good! You're welcome! :)

Answer 28

What did you get for the final answer? (?,?,?)

Answer 29

@mathmate
I got (1,1,0)
I'm about to test it out to see if I'm right...

Answer 30

Yep, it's the correct answer! Yaaay! Thank you so much for helping! This was giving me such a headache... phew! Glad it's finally solved! :)

Answer 31

hey could someone help me with my math cause i'm really confused and i need someone to explain

Answer 32

@athalia
I'm sure somebody will help you... good luck with it! It looks too confusing for me! Sorry!

Answer 33

wow ok thxs anyway

Answer 34

@dinorap1 Good, that was the correct answer, guess you have already checked!