Carissa15
  • Carissa15
please help, I have been unable to make any sense from this set of linear equations. Does this mean there is no solution? x-2y+z=0 2y-8z=8 -4x+5y+9z=-9
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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texaschic101
  • texaschic101
x - 2y + z = 0 --- multiply by 4 -4x + 5y + 9z = -9 ---------------- 4x - 8y + 4z = 0 (result of multiplying by 4) -4x + 5y + 9z = -9 ----------------add -3y + 13z = -9 -3y + 13z = -9 --- multiply by 2 2y - 8z = 8 -- multiply by 3 -------------- -6y + 26z = -18 (result of multiplying by 2) 6y - 24z = 24 (result of multiplying by 3) -------------add 2z = 6 z = 6/2 z = 3 2y - 8z = 8 2y - 8(3) = 8 2y - 24 = 8 2y = 24 + 8 2y = 32 y = 32/2 y = 16 x - 2y + z = 0 x - 2(16) + 3 = 0 x - 32 + 3 = 0 x - 29 = 0 x = 29 now we will check... 2y - 8z = 8 2(16) - 8(3) = 8 32 - 24 = 8 32 = 32 (correct) check... x - 2y + z = 0 29 - 2(16) + 3 = 0 29 - 32 + 3 = 0 32 - 32 = 0 0 = 0 (correct) so x = 29, y = 16, and z = 3
texaschic101
  • texaschic101
@Carissa15 .....check it out....any questions ?
texaschic101
  • texaschic101
I took my 1st and 3rd equation and eliminated the x. I then took my 2nd equation and the results of my first set of equations, and eliminated the y's....thus finding z. I then subbed z into the 2nd equation and found y. Subbed that into original equation and found x. I then checked my answers and they came out true.

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texaschic101
  • texaschic101
not really that hard.....just very,very time consuming
Carissa15
  • Carissa15
great. Makes perfect sense. Will work it out on paper myself to make sure I can get the same. Can I check a couple more with you please?
texaschic101
  • texaschic101
sure...what ya got ?
Carissa15
  • Carissa15
x-2y+z=0 2y-8z=8 -4x+5y+9z=-9 x+y+z=1 This is throwing me as there are four equations.
texaschic101
  • texaschic101
wow...I have never done it with 4....let me see what I can do...this could take awhile.
Carissa15
  • Carissa15
My textbook only covers 2 and some simple 3..
Carissa15
  • Carissa15
Could the same values from the last equation work?
texaschic101
  • texaschic101
x - 2y + z = 0 --- multiply by 4 -4x + 5y + 9z = -9 ------------- 4x - 8y + 4z = 0 (result of multiplying by 4) -4x + 5y + 9z = -9 --------------add -3y + 13z = -9 -3y + 13z = -9 --- multiply by 2 2y - 8z = 8 -- multiply by 3 ------------- -6y + 26z = -18 (result of multiplying by 2) 6y - 24z = 24 (result of multiplying by 3) ------------- 2z = 6 z = 6/2 z = 3 2y - 8z = 8 2y - 8(3) = 8 2y - 24 = 8 2y = 24 + 8 2y = 32 y = 32/2 y = 16 x - 2y + z = 0 x - 2(16) + 3 = 0 x - 32 + 3 = 0 x - 29 = 0 x = 29 check... 2y - 8z = 8 x - 2y + z = 0 2(16) - 8(3) = 8 29 - 2(16) + 3 = 0 32 - 24 = 8 29 - 32 + 3 = 0 8 = 8 (correct) 0 = 0 (correct) -4x + 5y + 9z = -9 x + y + z = 1 -4(29) + 5(16) + 9(3) = -9 29 + 16 + 3 = 1 -116 + 80 + 27 = -9 48 = 1 (oops....not correct) -116 + 107 = -9 -9 = -9 (correct) I have run into a problem.....for 3 of the equations, my answers are true...but for 1 of them, it is not true. Is x + y + z = 0 written correctly ??
texaschic101
  • texaschic101
oops..Is x + y + z = 1 written correctly ??
texaschic101
  • texaschic101
weird....I am getting the same answer as the last question
Carissa15
  • Carissa15
yes it is x+y+z=1
Carissa15
  • Carissa15
the first 3 equations are the same as the previous and the 4th is an added equation.
Carissa15
  • Carissa15
but the new equation means that for this question the 3 equations do not work as 48 does not equal 1?
texaschic101
  • texaschic101
I am not sure about this one....I am so sorry....I think we are gonna have to tag someone
texaschic101
  • texaschic101
sweet....john is here :)
johnweldon1993
  • johnweldon1993
No you got it right @texaschic101 I get no solutions for this :)
texaschic101
  • texaschic101
my brain is fried right now...lol.
Carissa15
  • Carissa15
Thank you so much.
johnweldon1993
  • johnweldon1993
I give you credit, that was the first time I've seen a system of 4 equations O.o haha but awesome work!
texaschic101
  • texaschic101
yep....thanks john :)
Carissa15
  • Carissa15
I also have this one I am stuck on (due to there not being a number on the right hand side to add/subtract from)
Carissa15
  • Carissa15
\[2x-y+3z=\alpha \] \[3x+y-5z=\beta \] \[-5x-5y+21z=\gamma \]
johnweldon1993
  • johnweldon1993
Oh that's interesting Right off the bat though I want to say, you cannot solve a system of equations where you have more variables than equations And here I count 3 equations but a total of 6 variables O.o impossible :)
Carissa15
  • Carissa15
that is what isn't making sense to me.
Carissa15
  • Carissa15
hmm, i just found a footnote.... show that the system is inconsistent if
johnweldon1993
  • johnweldon1993
Ahh, lol there ya go
Carissa15
  • Carissa15
\[\gamma \neq 2\alpha-3\beta \]
Carissa15
  • Carissa15
oops..
Carissa15
  • Carissa15
hmm, not seeming to get far with this. Any tips?
johnweldon1993
  • johnweldon1993
Oh sorry, stepped away for a bit Hmm, well lets see \[\large 2x - y + 3z = \alpha\] \[\large 3x + y - 5z = \beta\] \[\large -5x - 5y + 21z = \gamma\] And \(\large \gamma \cancel{=} 2\alpha - 3\beta\)
johnweldon1993
  • johnweldon1993
Well if we multiply the first equation by 2 And the second equation by -3 And then add them...what do we get?
johnweldon1993
  • johnweldon1993
Reasoning why I'm doing that, is because it is known we somehow need 2alpha...and we somehow need -3beta...so do that right off the bat And it turns out simpler than you think :D
Carissa15
  • Carissa15
Thank you @johnweldon1993 sorry, my computer crashed last night. Thank you for your help. Makes much more sense now :)

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