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Carissa15
 one year ago
please help, I have been unable to make any sense from this set of linear equations. Does this mean there is no solution?
x2y+z=0
2y8z=8
4x+5y+9z=9
Carissa15
 one year ago
please help, I have been unable to make any sense from this set of linear equations. Does this mean there is no solution? x2y+z=0 2y8z=8 4x+5y+9z=9

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x  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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Carissa15 .....check it out....any questions ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not really that hard.....just very,very time consuming

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0great. 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sure...what ya got ?

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0x2y+z=0 2y8z=8 4x+5y+9z=9 x+y+z=1 This is throwing me as there are four equations.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wow...I have never done it with 4....let me see what I can do...this could take awhile.

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0My textbook only covers 2 and some simple 3..

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0Could the same values from the last equation work?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x  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 ??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops..Is x + y + z = 1 written correctly ??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0weird....I am getting the same answer as the last question

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0the first 3 equations are the same as the previous and the 4th is an added equation.

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0but the new equation means that for this question the 3 equations do not work as 48 does not equal 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am not sure about this one....I am so sorry....I think we are gonna have to tag someone

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sweet....john is here :)

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1No you got it right @texaschic101 I get no solutions for this :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my brain is fried right now...lol.

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1I give you credit, that was the first time I've seen a system of 4 equations O.o haha but awesome work!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep....thanks john :)

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0\[2xy+3z=\alpha \] \[3x+y5z=\beta \] \[5x5y+21z=\gamma \]

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0that is what isn't making sense to me.

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0hmm, i just found a footnote.... show that the system is inconsistent if

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Ahh, lol there ya go

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0\[\gamma \neq 2\alpha3\beta \]

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0hmm, not seeming to get far with this. Any tips?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Oh 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
 one year ago
Best ResponseYou've already chosen the best response.1Well if we multiply the first equation by 2 And the second equation by 3 And then add them...what do we get?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Reasoning 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
 one year ago
Best ResponseYou've already chosen the best response.0Thank you @johnweldon1993 sorry, my computer crashed last night. Thank you for your help. Makes much more sense now :)
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