Question

Systems of Equations, please help!!

Answer 1

I know what the answers are but I cannot seem to get there =(

Answer 2

Let's focus on #24
\[\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2z = -3\\ z = 2x - 4y\\ \end{cases}\]

Answer 3

Notice we have a pair of 'z's here

\[\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2{\LARGE \color{red}{z}} = -3\\ {\LARGE \color{red}{z}} = 2x - 4y\\ \end{cases}\]

Answer 4

what we can do is replace the 'z' in the second equation with '2x-4y' since z = 2x-4y in the third equation

we can then drop the third equation after substitution


\[\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2\color{red}{z} = -3\\ \color{red}{z} = 2x - 4y\end{cases}\]
\[\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2\color{red}{(2x-4y)} = -3\end{cases}\]

Answer 5

For 24, which I think I maybe got right, I ended up with
\[y=3x-13, 4y-3x+2z=-3, z=2x-4y\]
and then substituted to end up with
\[4(3x-13)-3x+2(2x-4(3x-13))=-3\]
which simplified to
\[x=5, y=2, z=2\]

Answer 6

Is that correct (I have all the work but it's so long so I didn't post it sorry)

Answer 7

You did the right steps. Nice job

Answer 8

Awesome, thanks! Um do you think you could help with 25,26,27,28, or 29?

Answer 9

Those are definitely harder because one variable isn't already isolated. But we can isolate a variable, say z

if we pick on the third equation and solve for z, we get

3x - 2y - z = -9
3x - 2y - z+z = -9+z
3x - 2y = -9+z
3x - 2y + 9 = -9+z+9
3x - 2y + 9 = z
z = 3x - 2y + 9

Answer 10

agreed so far?

Answer 11

For which question?

Answer 12

#25

Answer 13

Okay =) Yeah that looks good

Answer 14

once we have z isolated, we can replace every copy of `z` in the first two equations with `(3x - 2y + 9)`

Answer 15

Okay, did that =)

Answer 16

first equation:
`x+3y-z = -4` will turn into `x+3y-(3x - 2y + 9) = -4`


second equation
`2x-y+2z=13` will turn into `2x-y+2(3x - 2y + 9) =13`

Answer 17

at this point, the 'z' has gone away leaving you with a system of 2 equations with 2 unknowns

Answer 18

That makes sense! I think I tried to overcomplicate it last time lol.
So from here, you would just isolate a variable in one of them and then plug whatever that is into the other equation?
Or would you solve both for a variable and then solve it like a regular system?

Answer 19

you can use a number of methods to solve this new system

1) substitution
2) elimination
3) matrices
4) graphing

graphing is probably the fastest way, but it doesn't always guarantee you get the exact answers (the point may have fractional coordinates)

Answer 20

I am most comfortable with substitution =)
So I solve both for a variable (let's say y, so y=something) and then just substitute (if y=x and y=2 then 2=x)?

Answer 21

you are correct

Answer 22

let me know what you get

Answer 23

Awesome thanks! Alright I'll do that real quick...

Answer 24

I got y=2x-13 and y=(-5/3)

Answer 25

x,y, & z are all whole numbers for #25

Answer 26

True. And the answer for this is x=0, y=1, z=7... let's see what I did wrong.

Answer 27

Ah. I combined x and y. I'll redo it

Answer 28

how did you get `y=2x-13`? which equation did you solve for y?

Answer 29

alright

Answer 30

I solved the first one to get that.

Answer 31

Here's what I did to solve the first eq.