Question

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

Answer 1

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

Answer 2

@Carissa15 .....check it out....any questions ?

Answer 3

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.

Answer 4

not really that hard.....just very,very time consuming

Answer 5

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?

Answer 6

sure...what ya got ?

Answer 7

x-2y+z=0
2y-8z=8
-4x+5y+9z=-9
x+y+z=1
This is throwing me as there are four equations.

Answer 8

wow...I have never done it with 4....let me see what I can do...this could take awhile.

Answer 9

My textbook only covers 2 and some simple 3..

Answer 10

Could the same values from the last equation work?

Answer 11

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 ??

Answer 12

oops..Is x + y + z = 1 written correctly ??

Answer 13

weird....I am getting the same answer as the last question

Answer 14

yes it is x+y+z=1

Answer 15

the first 3 equations are the same as the previous and the 4th is an added equation.

Answer 16

but the new equation means that for this question the 3 equations do not work as 48 does not equal 1?

Answer 17

I am not sure about this one....I am so sorry....I think we are gonna have to tag someone

Answer 18

sweet....john is here :)

Answer 19

No you got it right @texaschic101

I get no solutions for this :)

Answer 20

my brain is fried right now...lol.

Answer 21

Thank you so much.

Answer 22

I give you credit, that was the first time I've seen a system of 4 equations O.o haha but awesome work!

Answer 23

yep....thanks john :)

Answer 24

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)

Answer 25

\[2x-y+3z=\alpha \]
\[3x+y-5z=\beta \]
\[-5x-5y+21z=\gamma \]

Answer 26

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 :)

Answer 27

that is what isn't making sense to me.

Answer 28

hmm, i just found a footnote.... show that the system is inconsistent if

Answer 29

Ahh, lol there ya go

Answer 30

\[\gamma \neq 2\alpha-3\beta \]

Answer 31

oops..

Answer 32

hmm, not seeming to get far with this. Any tips?

Answer 33

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 \(\large \gamma \cancel{=} 2\alpha - 3\beta\)

Answer 34

Well if we multiply the first equation by 2
And the second equation by -3
And then add them...what do we get?

Answer 35

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

Answer 36

Thank you @johnweldon1993 sorry, my computer crashed last night. Thank you for your help. Makes much more sense now :)