Mathmatical Logic Fails Me

[BaronAsh]

Today for home work I was assigned a few extremely hard problems, One of these was:

Solve Each System Of Equations

2x-4y-z=10
4x-8y-2z=16
3x+y+z=12

____________________________________________


If I'm not mistaken it is logically impossible, as 2x-4y-z=10 multiplied by -2 is -4x+8y+2z=-20.

Now let's do some elimination.
-4x+8y+2z=-20
4x-8y-2z=16
______________
0=-4

(I'm sure -4 does not equal 0)

Thoughts and or help would be nice.

 

[Go Fighting In The War Room]

Could you show us more step in your process of elimination? That would help me try and figure this out quicker.

 

[A Weary Exile]

'No solution' perhaps? I'm not very good at math.

 

[BaronAsh]

Could you show us more step in your process of elimination? That would help me try and figure this out quicker.

-4x+4x=0
8y-8y=0
2z-2z=0
-20+16=-4 It's this last one that makes this problem impossible.

 

[wooty]

no help here at all, was reading that and the best i can come up with was to wiggle my finger over my lips and go "bibble bibble bibble"

thats considered schoolwork? good god

 

[Go Fighting In The War Room]

Could you show us more step in your process of elimination? That would help me try and figure this out quicker.

-4x+4x=0
8y-8y=0
2z-2z=0
-20+16=-4 It's this last one that makes this problem impossible.

Oh wow.... Maybe it's a trick. I have no answer for you. Sorry.

 

[dmase]

Shit i'm in precal 2 and i don't remember this shit i should be ashamed. Give me one sec need to get refreshed.

 

[Silver Scribbler]

Could you show us more step in your process of elimination? That would help me try and figure this out quicker.

-4x+4x=0
8y-8y=0
2z-2z=0
-20+16=-4 It's this last one that makes this problem impossible.

That is possible. -4 PLUS 20 = 16
so
16=16

 

[BlackJack47]

Sadly Math is not my strong suit. Science on the other hand

 

[Ultrasnail]

i think the top one is wrong otherwise you get 20=16.
if u change the first one it easily works.
id figure it out but im lazy and its your homework..... and its my birthday.

 

[Lukeje]

The determinant of the matrix equation is zero. Thus there is no solution.

 

[Headless Zombie]

What grade are you in? I was doing these about a month ago and we had plenty of impossible equations, we just wrote impossible if it was like yours. Only difference is we used matrices.

 

[BaronAsh]

You show 3 equations at the top, what happened to the third?

The third one doesn't really make a difference, if the first two have no solution.


To Handless Zombie I'm in 11th grade and we don't have matrices.

 

[tsb247]

I do not think there is a solution.

I plugged it into a 3x4 matrix and got it to reduced row echelon form. the numbers don't work.

This is what I get:

[[1 0 3/14 0]
[0 1 5/14 0]
[0 0 0 1]]

That is not proper reduced row echelon form - no solutions.

The proper form of a solvable equation would be something like:

[[1 0 0 x]
[0 1 0 y]
[0 0 1 z]]

So, mathematically, there is no solution for your system of equations.

Glad to be of service.

 

[hamster mk 4]

Yeah your sollution is the empty set.

Think about it this way.
A) 2x-4y-z=10
B) 4x-8y-2z=16
C) 3x+y+z=12

A, B, and C each represent a plane in 3 dimentional space. The question asks where in 3 dimentional space do these planes intersect. The problem is that A and B are parallel planes meaning they will never intersect. It is analigous to this problem:

y = 2x + 5
y = 2x + 7

Find the value of x and y that satisfies both equasions. You will see that two parralel lines will never cross. So the answer is the empty set or there is no answer.

If you wanted to get fancy you could calculate the lines where 3x+y+z=12 intersects 4x-8y-2z=16, and 3x+y+z=12 intersects 2x-4y-z=10. However there is no finite value of z, y, and z that satisfies all three equasions.

 

[BaronAsh]

-Snip-
Glad to be of service.

Thanks for the help.

 

[SomethingUnrelated]

The thing is, Maths is logic. It can't fail. If it did, somethings wrong, as in your case. Can't be arsed working it out though.

 

[Livinitlargeinabin]

Yeah, using Gaussian Elimination, it doesn't work

(2 -4 -1 | 10) => (2 -4 -1 | 10)
(4 -8 -2 | 16) => (0 0 0 | -4) - (R2=R2-2R1) <--- (This is impossible)
(3 1 1 | 12) => (0 14 5 | -6) - (R3=2R3-3R2)

The (0 0 0 | -4) means (0x + 0y + 0z = -4)
So as they say in my advanced higher maths (Scotland) class the system of equations is "Inconsistent"

Maybe thats the answer, the point might be to prove the fault.
So you're right, but check the question for your (+/-)s I always do stupid things like that.
Hope I've helped

 

[SnootyEnglishman]

I would try and help but i realized math sucks and i find that unless you plan on going into a type of job that requires knowing how to solve those types of equation then it's absolutely pointless.