## 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? x-2y+z=0 2y-8z=8 -4x+5y+9z=-9

1. 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

2. texaschic101

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

3. 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.

4. texaschic101

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

5. 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?

6. texaschic101

sure...what ya got ?

7. 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.

8. texaschic101

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

9. Carissa15

My textbook only covers 2 and some simple 3..

10. Carissa15

Could the same values from the last equation work?

11. 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 ??

12. texaschic101

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

13. texaschic101

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

14. Carissa15

yes it is x+y+z=1

15. Carissa15

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

16. Carissa15

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

17. texaschic101

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

18. texaschic101

sweet....john is here :)

19. johnweldon1993

No you got it right @texaschic101 I get no solutions for this :)

20. texaschic101

my brain is fried right now...lol.

21. Carissa15

Thank you so much.

22. johnweldon1993

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

23. texaschic101

yep....thanks john :)

24. 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)

25. Carissa15

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

26. 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 :)

27. Carissa15

that is what isn't making sense to me.

28. Carissa15

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

29. johnweldon1993

Ahh, lol there ya go

30. Carissa15

$\gamma \neq 2\alpha-3\beta$

31. Carissa15

oops..

32. Carissa15

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

33. 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 $$\large \gamma \cancel{=} 2\alpha - 3\beta$$

34. johnweldon1993

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

35. 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

36. Carissa15

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