Question

Can someone help me with these 3x3 Linear Systems
5x+2y+z=4
X+2z=4
2x+y-z=-1


3x+10y-12z
X-5y=0
X-4z=0

Answer 1

What is your favorite method of solution?

Answer 2

Elimination

Answer 3

Okay, what's preventing you from doing that?

x + 2z = 4 looks like a good place to start.

1) Multiply it by -5 and add to the first equation.
2) Multiply it by -2 and add it to the third equation.

You will have eliminated x from all but equation #2.

Answer 4

Teacher say try to use substitutions but i really dont like to use substitutions
I didnt realize that i can use elimination

Answer 5

Yes, elimination is just fine. I like to say, "Unique Answers don't care how you find them."

Answer 6

I can help but I use substitution best.

Answer 7

Is that alright @flowers ?

Answer 8

It should be pointed out that the first equation int eh second set is incomplete.

Answer 9

Sure
I need to learn it any way
:)

Answer 10

I haven't quite taken a look at the second set only the first one.

Answer 11

Alrighty :)

Due to the fact that the second equation `x+2z = 4` contains two variables we would change this into a function of x. To do this we would subtract 2z from each side.

\(\LARGE\bf{x+2z-2z=-2z+4 \rightarrow x = -2z + 4}\)

Answer 12

Now we would input this equation of x into the two other equations. We would simplify them.

\(\huge\bf{5(-2z+4)+2y+z=4}\)

What would this simplify to @flowers ?

Answer 13

Remember to first distribute into the parenthesis and then combine like terms.

Answer 14

Sorry open study likes to reshape it self
10 +20

Answer 15

\(\LARGE\bf{-10z+20+2y+z=4}\)

Combine like terms.

\(\LARGE\bf{-9z + 2y + 20 = 4}\)

Now we need the numbers without variables to be on one side. So we have to subtract 20 from both sides.

\(\LARGE\bf{-9z + 2y = 20 \color{red}{-20} = 4 \color{red}{-20}}\)

\(\huge\bf{-9z + 2y = -16}\)

Do you understand?

Answer 16

@flowers

Answer 17

So we leave this equation alone (we will use later). Now we input the equation of x into the 3rd equation.

\(\huge\bf{ 2(-2z+4)+y-z=-1}\)

What would this simplify to? @flowers

Answer 18

So sorry that equal sign near 20 is wrong its to be + not =. Darn the shift on my keyboard xD

Answer 19

4z+8

Answer 20

Nice :)

Now we combine like terms.

\(\LARGE\bf{-5z+y+8=-1}\)

Now the ones without a variable go to one side so we subtract 8 from both sides.

\(\LARGE\bf{-5z + y + 8 \color{red}{-8} = -1 \color{red}{-8}}\)

\(\LARGE\bf{-5z+y=-9}\)

Answer 21

Now we have.

\(\huge\bf{-9z+2y=-16}\)

and

\(\huge\bf{-5z + y = -9}\)

We will change the second equation into an equation of y since it is to itself. So we add 5z to each side.

\(\huge\bf{y = 5z -9}\)

~~~~~~~~~~~~~~~

Now we would input this equation of y into the first equation that we simplified.

\(\huge\bf{-9z + 2(\color{red}{5z-9})=-16}\)

What does that simplify to?

Answer 22

7z(-4z)=16

Answer 23

I think I'm way off

Answer 24

?

How did you come by 7z(-4z) = 16?

Remember when distributing you multiply with each term in the parenthesis.

\(\huge\bf{5z \times 2 -9 \times 2 = ?}\)

Answer 25

Dang it
Looking at the wrong thing

Answer 26

Funny I have the right idea on my paper but not here

Answer 27

Its fine :) Just make sure to keep your mind focus otherwise it will go off and problems will occur.

Answer 28

So we have.

\(\huge\bf{-9z + (5z \times 2 - 9 \times 2)=-16}\)

Answer 29

Woah
Multiple 5 and 2 right?

Answer 30

Yes :) First do the parenthesis.

Answer 31

But to be accurate `multiply 5z and 2`

Answer 32

OK
-9z+10z-18=-16

Answer 33

Nice :)

Now combine like terms.

\(\huge\bf{-9z + 10z - 18 = -16}\)

The like terms are `-9z and 10z`.

Answer 34

18z-18=-16

Answer 35

I mean 19z

Answer 36

Wait its a negative number
Ugh

Answer 37

Not quite.

\(\huge\bf{-9z + 10z = z}\)

Since we add -9 to a positive being 10 it will lower the number to 1.

So we have.

\(\huge\bf{z-18=-16}\)

Now we need z to itself so we add 18 to each side since it is being added to z.

\(\huge\bf{z-18 \color{red}{+18}=-16 \color{red}{+18}}\)

Answer 38

What does z equal?

Answer 39

Z=2

Answer 40

Yes :)

Answer 41

Seriously wow omg thank you so much

Answer 42

Now we need to input z into the equation that we turned into an equation of y `y = 5z -9`.

\(\huge\bf{y=5(2)-9}\)

What does y equal?

Answer 43

Not done yet ^.^ still need to find x and y.

Answer 44

3-9

Answer 45

Not quite.

\(\huge\bf{y=10-9}\)

Answer 46

Ik
Oops thinking differently
Meant to say 10

Answer 47

Thats alright ^.^ Though what is `10-9`?

Answer 48

1

Answer 49

Correct so y = 1 :) Now we need to find x.

Answer 50

Now we input y and z into the first equation `(5x + 2y + z = 4)`.

\(\huge\bf{5x+2(1)+2=4}\)

Simplify :)

Answer 51

Okay :)
5x+2+2=4

Answer 52

Nice :) Now we combine like terms `being 2 + 2`.

\(\huge\bf{5x+4=4}\)

Now we need `5x` to itself so we would subtract 4 from both sides.

\(\huge\bf{5x+4 \color{red}{-4} = 4 \color{red}{-4}}\)

What does that equal?

Answer 53

5x=0
/5 /5
X=0

Answer 54

Nice :)

Due to the fact it is formatted as \(\large\bf{(x,y,z)}\) your answer would be as such...

\(\huge\bf\color{red}{(0,1,2)}\)

Answer 55

Yay thank you so much
I can now do the other problems
;)

Answer 56

\(\color{#0cbb34}{\text{Originally Posted by}}\) @tkhunny
It should be pointed out that the first equation in the second set is incomplete.
\(\color{#0cbb34}{\text{End of Quote}}\)

I can not really help for the second set due to the fact the first is incomplete as hunny said. `"3x+10y-12z"` it seems to be missing its total.

Answer 57

Ikr but this algebra 2 so yea..

Answer 58

Are you sure its only...

`3x + 1y - 12z`

in order to solve 3x3 systems you need to have a total in each equation. Are you sure its not missing something. `3x + 1y - 12z = ?`

Answer 59

Let me check

Answer 60

Oops 3x+10y-12z=40

Answer 61

Ooo ok good to know. I know this is bad timing but I must take my leave for the afternoon. @tkhunny can you help @flowers ?

Answer 62

or @Directrix @zepdrix @mathmate ?

Answer 63

Oh no
Your fine
I can do it

Answer 64

Thank you for your help

Answer 65

Oh ok. Well good luck with your work and I am glad I was able to help ^.^ See ya around.

Answer 66

Your very welcome :)