/sci/ - Science & Math

[Anonymous No.2756179]

first time on /sci/ so be nice....

can anyone explain or at least solve this system of equations for me?

x + y + z = 3
x - y + z = 3
2x + y + z = 4

I know its simple but I need help

[Anonymous No.2756185]

This is not /homework help/

[Anonymous No.2756193]

>>2756185
>i don't know how to do it so i'm just going to make a snide comment

[Anonymous No.2756200]

OP this is basic algebra, if you can't figure it out then you're either underage b& or you're fucking stupid. Probably both.

[Anonymous No.2756201]

x=1, y=0, z=2

[Anonymous No.2756207]

http://en.wikipedia.org/wiki/System_of_linear_equations#Solving_a_linear_system

[EK !EKFQOBUFnQ No.2756214]

x + y + z = 3
x - y + z = 3
x + y + z = x - y + z
x + y = x - y
y = - y
y = 0

2x + z = 4
x + z = 3
x = 1
z = 2

[bleedingedge !!OBNkFn9BVXZ No.2756218]

Use matricies!

[EK !EKFQOBUFnQ No.2756223]

>>2756214
dammit...

[Anonymous No.2756224]

You want to get rid of one of the variables.

Add (x + y + z = 3) and (x - y + z = 3). You have a new equation, 2x + 2z = 6.

Add (x - y + z = 3) and (2x + y + z = 4). Now another one, 3x + 2z = 7.

We still need to solve for a variable, let's get the variable x. This time we'll have to multiply either 2x + 2z = 6 or 3x + 2z = 7 by -1. Let's do the top one.

-2x - 2z = -6

Add them together now.

x = 1

Now, plug in 1 for x for your new equations.

2 + 2z = 6, 2z = 4, z = 2
3 + 2z = 7, 2z = 4, z = 2

Great, now plug in both z and x into your three original equations to get the variable y.

[zugz No.2756232]

x + y + z = 3 ----> x = 3 - y - z
x - y + z = 3 ----> (3 - y - z) - y + z = 3 ----> 3 - 2y = 3 ---> y = 0

so now we know x = 3 - z from the 1st line, and the 3rd eq is now 2x + z = 4
so 2(3 - z) + z = 4 .... and you're practically done

[Anonymous No.2756236]

The left side may be expressed as the matrix A:

\[ \left( \begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 1 \\
2 & 1 & 1 \end{array} \right)\]

And the solutions are similarly B:

\[ \left( \begin{array}{c}
3 \\
3 \\
4 \end{array} \right)\]

So A^{-1} B = C, your solution matrix, [ x y z ]

Alternatively, solve for a variable (let's say y) in the first, and plug into the second (reducing the second to two variables). Then solve for one (how about z), and plug into the last equation.

So now the last equation is only in terms of x. Work the arithmetic. and plug into the previous iteration that only included x and z. Solve for z and plug both values into the first original equation.

Also, my first time using TeX in /sci/, so if the code all goes to poo-poo, my apologies. They're the coefficient matrices of the system.

[Anonymous No.2756277]

OP here,

>>2756185 fuck you
>>2756193 thanks?
>>2756200 fuck you, nice dubs
>>2756201 that doesnt even help & its wrong
>>2756207 tl;dr
>>2756214 wut
>>2756218 fuck that, no
>>2756224 WINRAR

[EK !EKFQOBUFnQ No.2756316]

>>2756277
>that doesnt even help & its wrong
you rekon?

...why dont u try plugging those values into the 3 equations and see for yourself...

[Anonymous No.2756345]

>>2756277

>>2756224 here, I just explained a simple way to do it. You better learn to do it the right way.

[Anonymous No.2756400]

y=0

x=1

z=2

[Anonymous No.2756406]

>>2756214
Thanks EK, can you help me solve this now?

>>2755807

[Anonymous No.2756408]

this is why americunts suck at math. everything they know about it comes from 4chan

[Anonymous No.2756458]

<span class="math"> x+y+z=3 \Rightarrow x=3-y-z [/spoiler] Solve for x.

<span class="math"> x-y+z=3 \Rightarrow (3-y-z)-y+z=3 \Rightarrow -2y=0 \Rightarrow y=0 [/spoiler] Substituting for x and solving for y.

<span class="math"> 2x+y+z=4 \Rightarrow 2(3-0-z)+0+z=4 \Rightarrow 6-2z+z=4 \Rightarrow -z=-2 \Rightarrow z=2 [/spoiler] Substituting in x and y then solving for z.

<span class="math"> x+y+z=3 \Rightarrow x+0+2=3 \Rightarrow x=1 [/spoiler] Plugging in y and z to solve for x.

There you go, <span class="math"> (1,0,2) [/spoiler]