Question

5x+7y=-1
-2y+3z=9
7x-z=27

So I got a bit lost on this problem. Can someone walk me through?

Answer 1

what methods do you know?

Do you know how to use
matrix
substitution
add or subtract the equations from each other

Answer 2

Uhm for this one its either substitution or elimination

Answer 3

No matrix

Answer 4

Ok, I have a thought...

You can solve for z in the last equation, and then solve for y in the first equation and then you should be able to solve for x by substituting into the second equation.

Answer 5

you can use matrix too, if you know how to. That's what I would suggest.

Answer 6

My teacher said not to do matrix plus I haven't learnt it yet from her. We learn that tomorrow. But I have to know how to do this cuz I have a quiz tomorrow

Answer 7

Oh, ok.

Answer 8

Lemme tell yall how she told me to go about with the steps of solving.

Answer 9

Then do substitution,

Answer 10

5x+7y=-1
-2y+3z=9
7x-z=27

(recalling)

Answer 11

1)pick a variable to eliminate
2) pick a pair of equations and eliminate that variable
3)pick a different pair of equations and eliminate the same variable.
4)solve the two resulting equations

Answer 12

Yup

Answer 13

eliminate z in the last equation and substitute into the second one.

Answer 14

Do I not pick a pair twice tho?

Answer 15

you will result with 2 variables after you substitute the z from the last equation to the second

Answer 16

Like, I did elimination and took out the y in the 1st and 2nd equation, then in the 2nd and 3rd equation I eliminated the y.
Then I was confused where to go next

Answer 17

yes correct, I resulted in 2 variables

Answer 18

tell me what exactly what you get, write it out.

Answer 19

Ok.

Answer 20

I mean when you get 2 variables....

Answer 21

\(~~~~5x+7y=-1\\
-2y+3z=9\\
~~~~~~7x-z=27\)

We have this correct?

Well we can take the third line, and solve for z.

\(7x-z=27\)
\(-z=-7x+27\)
\(z=7x-27\)

Then we can take the first line and solve for y.

\(5x+7y=-1\)
\(7y=-5x-1\)
\(\displaystyle y=-\frac{5}{7}x-\frac{1}{7}\)

Then we should be able to substitute it back into the second equation.

\(-2y+3z=9\)

\(\displaystyle -2(-\frac{5}{7}x-\frac{1}{7})+3(7x-27)=9\)

Now we have all x's in this line and can therefore solve for x right?

After you have that you should be able to easily solve out for the remaining variables.

Answer 22

@austinL, why are you doing his work for him?

Answer 23

It was significant enough that I deemed it more appropriate to do it this way so that he can see it uninterrupted.

Answer 24

1st step

5x+7y=-1
-2y+3z=9

10x+14y=-2
+
21z-14y=63
-------------
10x+21z=61

2nd step
-2y+3z=9
7x-z=27

-2y+3z=9
+
21x-3z=81
-----------
-2y+21x=90

Answer 25

Those are the steps I have gone through so far..now I am confused where to take those two equations from there.

Answer 26

let me check....

Answer 27

The next step is to solve the resulting equations...how to do that?!

Answer 28

The thing is, there are 3 variables but only 2 variables in each line making elimination method difficult to use.

Answer 29

My bad it is -2

Answer 30

-2+63=61 lol

Answer 31

Yeah, you are right!

Answer 32

So then how do I use substitution after my steps? @austinL

Answer 33

alright where do I take it from there then? @SolomonZelman

Answer 34

I keep losing the equations b/c of the replies, Ill right them on my paper...

Answer 35

Alright

Answer 36

Go back and look at my steps, that is how I would set about solving this problem.

Answer 37

I see you do substitution before, Yeah I need to learn it my teachers way..

Answer 38

look at your first equation (of the initial once, that are in your question)

Answer 39

5x+7y=-1

Answer 40

ok.
and the one that you got
-2y+21x=90

Answer 41

ok.

Answer 42

( I agree that substitution is easier though...)

Answer 43

Wait, what about me other equation!!!

Answer 44

But I fail to see how this(your teachers method) can work with this problem. It works just dandy for systems set up like.

ax+by+cz=d
ex+fy+gz=h
ix+jy+kz=l

yours is set up,

ax+by=c
-dy+ez=f
gx-hz=i

Answer 45

Right, I agree. Somehow she made it work tho..

Answer 46

You said this is for a quiz tomorrow right? Does the REQUIRE you to use the method you mentioned above?

Answer 47

it probably would, I like substitution and matrix better....

Answer 48

karatechopper, it would work better with 4 equations, but 3 equations like these ones, Nooooooooooo

Answer 49

Yes. it requires it

Answer 50

Normally you subtract one equation from both other equations to get rid of a variable completely...

Answer 51

\( 5x+7y=-1\)
\(-2y+3z=9\)
\(7x-z=27\)

This is what we have.

Lets take the first two lines and eliminate y,

\(5x+7y=-1\)
\(-2y+3z=9\)

\(2(5x+7y=-1)\)
\(7(-2y+3z=9)\)

This will give,

\(10x+14y=-2\)
\(\underline{-14y+21z=63}\)
\(10x+21z=61\)

Then we can take that and the last line,

\(10x+21z=61\)
\(7x-z=27\)

You can then eliminate one variable and solve for one of them. You can then substitute back in what you have easily to solve for the remaining variables.

Answer 52

Anywho, I have to go now, take care guys!

Answer 53

oh..i see..

Answer 54

@austinL is right! Can you do it from there?

Answer 55

yes you could.......

Answer 56

I believe so yes!

Answer 57

are you ready for a quiz?

Answer 58

Not really. I need for practice on these problems.

Answer 59

Would you give me a quiz? I need to practice problems like this one. I think I am good with the 3 variables in each equation one