anonymous
  • anonymous
Systems of Equations, please help!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I know what the answers are but I cannot seem to get there =(
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jim_thompson5910
  • jim_thompson5910
Let's focus on #24 \[\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2z = -3\\ z = 2x - 4y\\ \end{cases}\]
jim_thompson5910
  • jim_thompson5910
Notice we have a pair of 'z's here \[\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2{\LARGE \color{red}{z}} = -3\\ {\LARGE \color{red}{z}} = 2x - 4y\\ \end{cases}\]

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jim_thompson5910
  • jim_thompson5910
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 \[\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2\color{red}{z} = -3\\ \color{red}{z} = 2x - 4y\end{cases}\] \[\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2\color{red}{(2x-4y)} = -3\end{cases}\]
anonymous
  • anonymous
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 \[x=5, y=2, z=2\]
anonymous
  • anonymous
Is that correct (I have all the work but it's so long so I didn't post it sorry)
jim_thompson5910
  • jim_thompson5910
You did the right steps. Nice job
anonymous
  • anonymous
Awesome, thanks! Um do you think you could help with 25,26,27,28, or 29?
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
agreed so far?
anonymous
  • anonymous
For which question?
jim_thompson5910
  • jim_thompson5910
#25
anonymous
  • anonymous
Okay =) Yeah that looks good
jim_thompson5910
  • jim_thompson5910
once we have z isolated, we can replace every copy of `z` in the first two equations with `(3x - 2y + 9)`
anonymous
  • anonymous
Okay, did that =)
jim_thompson5910
  • jim_thompson5910
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`
jim_thompson5910
  • jim_thompson5910
at this point, the 'z' has gone away leaving you with a system of 2 equations with 2 unknowns
anonymous
  • anonymous
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?
jim_thompson5910
  • jim_thompson5910
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)
anonymous
  • anonymous
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)?
jim_thompson5910
  • jim_thompson5910
you are correct
jim_thompson5910
  • jim_thompson5910
let me know what you get
anonymous
  • anonymous
Awesome thanks! Alright I'll do that real quick...
anonymous
  • anonymous
I got y=2x-13 and y=(-5/3)
jim_thompson5910
  • jim_thompson5910
x,y, & z are all whole numbers for #25
anonymous
  • anonymous
True. And the answer for this is x=0, y=1, z=7... let's see what I did wrong.
anonymous
  • anonymous
Ah. I combined x and y. I'll redo it
jim_thompson5910
  • jim_thompson5910
how did you get `y=2x-13`? which equation did you solve for y?
jim_thompson5910
  • jim_thompson5910
alright
anonymous
  • anonymous
I solved the first one to get that.
anonymous
  • anonymous
|dw:1442627727801:dw| Here's what I did to solve the first eq.
jim_thompson5910
  • jim_thompson5910
You made a mistake in distributing x+3y-(3x - 2y + 9) = -4 turns into x+3y-3x + 2y - 9 = -4
jim_thompson5910
  • jim_thompson5910
notice the `+ 2y` and ` - 9`
anonymous
  • anonymous
OH! I was wondering about that. So the negative sign gets distributed!
jim_thompson5910
  • jim_thompson5910
yeah think of `-(3x - 2y + 9)` as `-1*(3x - 2y + 9)`
jim_thompson5910
  • jim_thompson5910
and that -1 multiplies with each term inside
anonymous
  • anonymous
All right =) Here, I'll try that again
anonymous
  • anonymous
I got y=(2/5x)+1
anonymous
  • anonymous
and y=(8/5x)+1 for the other which when set equal to eachother make x=0 which is correct
jim_thompson5910
  • jim_thompson5910
I'm getting the same equations when I solve for y
anonymous
  • anonymous
Yay! When I plugged x=0 into the other eq.'s, it worked! Thank you so much!
jim_thompson5910
  • jim_thompson5910
no problem
anonymous
  • anonymous
I think I should be able to do the rest now. Thanks again =)
jim_thompson5910
  • jim_thompson5910
sure thing

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