#5**+15 **

SHHHHHHHHH

Nobody knows ( that final substitution, i typed -2 instead of -4, no biggy)

SpawnofAngel
Apr 6, 2016

#1**+13 **

Solve the following system:

{x+2 y+4 z = 3 | (equation 1)

x-y-2 z = -3 | (equation 2)

3 x+z = 0 | (equation 3)

Swap equation 1 with equation 3:

{3 x+0 y+z = 0 | (equation 1)

x-y-2 z = -3 | (equation 2)

x+2 y+4 z = 3 | (equation 3)

Subtract 1/3 × (equation 1) from equation 2:

{3 x+0 y+z = 0 | (equation 1)

0 x-y-(7 z)/3 = -3 | (equation 2)

x+2 y+4 z = 3 | (equation 3)

Multiply equation 2 by -3:

{3 x+0 y+z = 0 | (equation 1)

0 x+3 y+7 z = 9 | (equation 2)

x+2 y+4 z = 3 | (equation 3)

Subtract 1/3 × (equation 1) from equation 3:

{3 x+0 y+z = 0 | (equation 1)

0 x+3 y+7 z = 9 | (equation 2)

0 x+2 y+(11 z)/3 = 3 | (equation 3)

Multiply equation 3 by 3:

{3 x+0 y+z = 0 | (equation 1)

0 x+3 y+7 z = 9 | (equation 2)

0 x+6 y+11 z = 9 | (equation 3)

Swap equation 2 with equation 3:

{3 x+0 y+z = 0 | (equation 1)

0 x+6 y+11 z = 9 | (equation 2)

0 x+3 y+7 z = 9 | (equation 3)

Subtract 1/2 × (equation 2) from equation 3:

{3 x+0 y+z = 0 | (equation 1)

0 x+6 y+11 z = 9 | (equation 2)

0 x+0 y+(3 z)/2 = 9/2 | (equation 3)

Multiply equation 3 by 2/3:

{3 x+0 y+z = 0 | (equation 1)

0 x+6 y+11 z = 9 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Subtract 11 × (equation 3) from equation 2:

{3 x+0 y+z = 0 | (equation 1)

0 x+6 y+0 z = -24 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Divide equation 2 by 6:

{3 x+0 y+z = 0 | (equation 1)

0 x+y+0 z = -4 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Subtract equation 3 from equation 1:

{3 x+0 y+0 z = -3 | (equation 1)

0 x+y+0 z = -4 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Divide equation 1 by 3:

{x+0 y+0 z = -1 | (equation 1)

0 x+y+0 z = -4 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Collect results:

**Answer: | x = -1 y = -4 z = 3**

Guest Apr 6, 2016

#2**+10 **

elimination for first 2

Apply a negative to the second one to eliminate X

x+2y+4z = 3

-x+y+2z = 3

0x+3y+6z = 6

We can take a 3 out of everything

y+2z=2

y = 2-2z

Likewise we can make z = -3x with the last equation

So we have

y= 2-2z

z = -3x

Which means y = 2+6x

Lets substitute for the second equation

x - (2+6x) - 2(-3x) = -3

Solve for x

It boils down to

x-2=-3

So x = -1

Substitute into original equations

3(-1) + z=0

z = 3

Second equation would be easiest for the last substitution

-1 - y - 2(3) = -3

y = -4

x = -1

y = -4

z = 3

SpawnofAngel
Apr 6, 2016

#3**+10 **

You've slipped up somewhere Spawn. Guest#1 has the right results (albeit with a rather long solution method!)

Alan
Apr 6, 2016

#5**+15 **

Best Answer

SHHHHHHHHH

Nobody knows ( that final substitution, i typed -2 instead of -4, no biggy)

SpawnofAngel
Apr 6, 2016

#6**+5 **

Here's another approach.....

x + 2y + 4z = 3 (1)

x - y - 2z = -3 (2)

3x + z = 0 → z = -3x (3)

Substitute (3) into (2)

x - y - 2(-3x) = -3

x - y + 6x = -3

7x - y = -3

y = 7x + 3 (4)

Substitute (3) and (4) into (1)

x + 2 [7x + 3] + 4(-3x) = 3

x + 14x + 6 - 12x = 3

3x = -3 x = -1

Using (4), y = 7(-1) + 3 = -4

Using (3), z = -3(-1) = 3

So {x, y, z } = { -1 , -4, 3 }

CPhill
Apr 6, 2016