Problem with linear equation

[ZOGs #1 Supporter]

7x+5y-z=28
x-4y-z=5
4x-5y-2z=19

Feel free to use any technique to solve. I'm using simple elimination and substitution to solve as I got to refresh my mind on this.

 

[Stephen]

7x+5y-z=28
-(x-4y-z=5)
=6x+9y=23
(x-4y-z=5)(2)
-(4x-5y-2z=19)
=-2x-3y=-9
6x+9y=23
+(-2x-3y=-9)(3)
=KILL YOURSELF
7x+5y-z=28
-(x-4y-z=5)(7)
=33y+6z=-7
(x-4y-z=5)(4)
-(4x-5y-2z=19)
=11y-2z=1
33y+6z=-7
-(11y-2z=1)(3)
=NOOOOOOOOOO
(7x+5y-z=28)(4/5)
-(x-4y-z=5)(5/4)
28x/5-4z/5=112/4
-(5x/4-5z/4=25/4)
23x I forgot this part of math I give up

If I completely did the process wrong, I do not apologize nor regret it

 

[princessparkles]

i think your question just sucks

 

[ZOGs #1 Supporter]

i think your question just sucks View attachment 21285

I thought it would be like this as I kept getting 2x+3y equalling to different values. oh BTW in grade 12 calculus and vectors, an equation like the above is how you do planes in 3-d. Also, silly woman, when a variable does not have any discernible value, that variable has a little something known as 'no solutions'. A system of linear equations not having any solutions is a feature not a buggy.

 

[Stephen]

I forgot to put in a final answer to the problem... it_hurts

 

[ZOGs #1 Supporter]

I forgot to put in a final answer to the problem... it_hurts

 

[footjob]

do you need help on this or are you just quizzing us i hope its the latter

 

[footjob]

nevermind its an impossible question award

 

[ZOGs #1 Supporter]

why are you niggers so stupid the answer to the question is there are no answers.

 

[itransheartsoy]

7x+5y-z=28
x-4y-z=5
4x-5y-2z=19

Feel free to use any technique to solve. I'm using simple elimination and substitution to solve as I got to refresh my mind on this.

 

[(In)CelibatePsycho]

I actually forgot how to do these by hand because I would always just use my calculator since it’s faster and easier