Question

Can someone please help me solve a system using elimination? its for algebra 2.

Answer 1

yeah okay

Answer 2

give it ot me

Answer 3

Okay one sec

Answer 4

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

Answer 5

@518nad

Answer 6

you can add some multiples of any two equations together and have a variable drop out, say rewriting equation 3, as equation3+equation1

Answer 7

I tried to do that but there were so many steps and I got confused @DanJS

Answer 8

I've gotten a different answer every time

Answer 9

Here it is after doing eq3=eq3+eq1, and eq2=(-1)*eq2+eq1
-2x+2y+3z=0
3y+2z=3
5y+6z=5
------------------

Answer 10

So those are the same equations,
just multiplied by -1?

Answer 11

yeah, i multiplied equation 2 by -1, then added equation 1 to it

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

Answer 12

Sorry I'm still kinda lost... the first equation you wrote is equation one, correct?

Answer 13

And how did you get the next two equations?

Answer 14

ok, when you have a system of linear equations like that, you can do 3 things.

1 - change the order of the equations
2 - multiply an equation by a constant number
3 - add equations together

Answer 15

Okay

Answer 16

So which one is the best option?

Answer 17

So starting from the given, you want to add equations together so that one of the variables goes to zero and drops out.

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

Rewrite equation3, as equation3+equation 1, so that the -2x+2x, will go to 0x and x will dissapear, leaving an equation in only y and z.

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

good on that so far?

Answer 18

I understand the first part, but how did you get 5y+6z=5 ?

Answer 19

that is the new 3rd equation. it is now eq1+eq3, adding equation 1 to equation 3 gives that

Answer 20

Okay I understand that part now

Answer 21

now do the same sorta thing to eliminate the x variable from either eq1 or eq 2

Answer 22

Okay one sec

Answer 23

So if I multiply eq1 by -1 then add that to eq2, it will cancel out the x value?

Answer 24

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

you see you need to multiply one of them by -1 first, so the x terms add up to zero

-1 times equation 2 gives
-2x+2y+3z=0
2x+y-z=3
5y+6z=5
---------------

and then do equation 2 goes to the sum equation 2 plus equation 1, Eq2--->Eq2 + Eq1

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

Answer 25

yeah you got it

Answer 26

Okay I'm just reading it all one sec

Answer 27

You just have to remember what you can do, those 3 rules, and try then to use those to eliminate variables from equations.

Answer 28

Okay thanks for all the help

Answer 29

You can do the same sort of thing to get rid of y or z in eq2 or eq3. This will leave an equation with 1 variable.

Answer 30

-3 times equation 2

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

Eq3 becomes eq3+eq2

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

Answer 31

Now you have enough to solve for each variable, the third equation solves to y=1, then use that to get the x and z values

Answer 32

Okay I will try

Answer 33

So would y=1? @DanJS

Answer 34

yes, from equation 3 , -4y=-4, y=1

Answer 35

use y=1 in equation 2 now, to see that z will be zero

3y+2z=3 ----- 3*1 + 2z = 3

z=0

Answer 36

now use those two y=1 and z=0, in equation 1 to get x

-2x+2y+3z=0
-2x+2*1+3*0=0
-2x=-2
x=1

x also is 1, so the three variables are

x=1
y=1
z=o

you can test by putting those values into each equation, all 3 equations should hold true

Answer 37

Okay, I understand now. Thank you so much it was very helpful. @DanJS

Answer 38

welcome