k, lets see if you guys could answer some good ol' Algebra 2 questions.

[5 2 -3] [a] [-5]
[-4 0 -4] [b] = [20]
[-3 -5 -5] [c] [53]

Solve for A, B, and C.

Note: This is only one problem, so each line doesn't seperate a different problem. You are trying to figure out how the first column of 3 numbers multiplied by the column ABC will equal the last column of 3 numbers.

[quote="[DAS REICH] Blitz":69893]k, lets see if you guys could answer some good ol' Algebra 2 questions.

[5 2 -3] [a] [-5]
[-4 0 -4] [b] = [20]
[-3 -5 -5] [c] [53]

Solve for A, B, and C.

Note: This is only one problem, so each line doesn't seperate a different problem. You are trying to figure out how the first column of 3 numbers multiplied by the column ABC will equal the last column of 3 numbers.[/quote:69893]

Here's one for you....

Blitz + check PM's = ?

stay in school kids

[quote="9mm BeRetTa":d3d1a]just make all the denominators the same and its easy oOo:

( 3y - 1y = 20)
(2y = 20)
(y=10)[/quote:d3d1a]

i did the same thing

[code:2f1eb]

X X Y Z ?

X W Y W 85

Z Y X Z 87

W X Y Z 82

87 86 93 79
[/code:2f1eb]

Solve the first line. There are two methods one can use to arrive at the
correct answer and unfortunately, the easy way isn't, by reason of four
variables. Let's see who knows his Algebra...

Minor edit:

1) All lines are composed exclusively of additions. (W+X+Y+Z=82)
2) I will post the complete solution in a day.

^ i dont

[quote="[DAS REICH] Blitz":cda00]
[5 2 -3] [a] [-5]
[-4 0 -4] [b] = [20]
[-3 -5 -5] [c] [53]

Solve for A, B, and C.
[/quote:cda00]

I've understood the above to mean the equivalent of

5a + 2b - 3c = -5
-4a - 4c = 20
-3a - 5b - 5c = 53

after manipulation of the matrices, in which case the problem solves for
A = -1
B = -6
C = -4

I can post the entire solution if you want.

Here's an Algebra problem I've created:

[code:53b64]

X X Y 40

Z Y Z 50

X Y Z ?

35 50 50
[/code:53b64]

I want you to solve it WITHOUT resorting to an individual variable solution,
that is to say I don't want you solving X, then Y and Z: there's an alternate
way of finding the value of the line. Simple numerical answers will not count:
I want the details of your solution.

[quote="Mr.Buttocks":e2813][quote="[DAS REICH] Blitz":e2813]k, lets see if you guys could answer some good ol' Algebra 2 questions.

[5 2 -3] [a] [-5]
[-4 0 -4] [b] = [20]
[-3 -5 -5] [c] [53]

Solve for A, B, and C.

Note: This is only one problem, so each line doesn't seperate a different problem. You are trying to figure out how the first column of 3 numbers multiplied by the column ABC will equal the last column of 3 numbers.[/quote:e2813]

Here's one for you....

Blitz + check PM's = ?[/quote:e2813]fin.

[quote=SoLiDUS]I've understood the above to mean the equivalent of

5a + 2b - 3c = -5
-4a - 4c = 20
-3a - 5b - 5c = 53

after manipulation of the matrices, in which case the problem solves for
A = -1
B = -6
C = -4

I can post the entire solution if you want.[/quote:7442e]

wow, you got it right... dam i thought no one would get it.

As promised, the solutions to my problems

[code:593bc]

X X Y 40

Z Y Z 50

X Y Z ?

35 50 50
[/code:593bc]

Solution:

(X+Y+Y) + (Y+Z+Z) + (X+Z+X) - (X+X+Y) - (Y+Z+Z) = X + Y + Z

( (X+Y+Y) + (Y+Z+Z) + (X+Z+X) ) - ( (X+X+Y) - (Y+Z+Z) ) = X + Y + Z

( 50 + 50 + 35 ) - ( 40 + 50 ) = 135 - 90 = 45


[code:593bc]

X X Y Z ?

X W Y W 85

Z Y X Z 87

W X Y Z 82

87 86 93 79
[/code:593bc]

Solution:

Use line worth 79 and solve w = 79 - 3z

2x + (79 - 3z) + z = 87
2x + 79 - 2z = 87
2x - 2z = 8
2x - 8 = 2z
Solve for z = x - 4

2 (x - 4) + x + y = 87
2x -8 + x + y = 87
3x + y = 95
Solve for y = 95 - 3x

3y + x = 93
3 (95 - 3x) + x = 93
285 - 8x = 93
192 = 8x
Solve for x = 24

Since z = x - 4,
Solve z = 24 - 4 = 20

Since y = 95 - 3x,
Solve y = 95 - 3(24) = 95 - 72 = 23

Since we know X, Y and Z, we can solve the first line:
24 + 24 + 23 + 20 = 91