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# Solving system of equations

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Apr 6, 2016

#5
+2592
+15

SHHHHHHHHH

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

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

Apr 6, 2016
#2
+2592
+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

Apr 6, 2016
edited by SpawnofAngel  Apr 6, 2016
#3
+28022
+10

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

Apr 6, 2016
edited by Alan  Apr 6, 2016
#4
+28022
+10

Ok, you've managed to correct it.

Alan  Apr 6, 2016
#5
+2592
+15

SHHHHHHHHH

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

SpawnofAngel  Apr 6, 2016
#6
+101369
+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 }

Apr 6, 2016