Math

2x+2y+z= -5      →   z = -2x - 2y - 5   (1)

3x+4y+2z=0   (2)

x+3y+2z=1     →   -x - 3y - 2z   = -1    (3)

Add (2) and (3)

2x + y  = -1   →  y  = -1 - 2x  (4)

Put  (1) and (4)  into (2)

3x + 4[ -1 - 2x] + 2 [ -2x - 2[-1 - 2x] -5]  =  0   simplify this

3x -4 - 8x + 2 [ -2x + 2 +4x - 5]  = 0

3x - 4 - 8x - 4x + 4 + 8x - 10  = 0

-x - 10  = 0

-10  = x

And y = -1- 2(-10)  = 19

And z = -2(-10) - 2(19) - 5 =   20 - 38  - 5  = -23

So

[x, y, z ]  = [ -10, 19, -23]

Explain how to solve the follwoing system of equations. What is the solution to the system?

2x+2y+z=-5

3x+4y+2z=0

x+3y+2z=1

Solve the following system:
{2 x+2 y+z = -5 |     (equation 1)
x+3 y+2 z = 1 |     (equation 2)
3 x+4 y+2 z = 0 |     (equation 3)
Swap equation 1 with equation 3:
{3 x+4 y+2 z = 0 |     (equation 1)
x+3 y+2 z = 1 |     (equation 2)
2 x+2 y+z = -5 |     (equation 3)
Subtract 1/3 × (equation 1) from equation 2:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+(5 y)/3+(4 z)/3 = 1 |     (equation 2)
2 x+2 y+z = -5 |     (equation 3)
Multiply equation 2 by 3:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+5 y+4 z = 3 |     (equation 2)
2 x+2 y+z = -5 |     (equation 3)
Subtract 2/3 × (equation 1) from equation 3:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+5 y+4 z = 3 |     (equation 2)
0 x-(2 y)/3-z/3 = -5 |     (equation 3)
Multiply equation 3 by -3:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+5 y+4 z = 3 |     (equation 2)
0 x+2 y+z = 15 |     (equation 3)
Subtract 2/5 × (equation 2) from equation 3:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+5 y+4 z = 3 |     (equation 2)
0 x+0 y-(3 z)/5 = 69/5 |     (equation 3)
Multiply equation 3 by 5/3:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+5 y+4 z = 3 |     (equation 2)
0 x+0 y-z = 23 |     (equation 3)
Multiply equation 3 by -1:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+5 y+4 z = 3 |     (equation 2)
0 x+0 y+z = -23 |     (equation 3)
Subtract 4 × (equation 3) from equation 2:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+5 y+0 z = 95 |     (equation 2)
0 x+0 y+z = -23 |     (equation 3)
Divide equation 2 by 5:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+y+0 z = 19 |     (equation 2)
0 x+0 y+z = -23 |     (equation 3)
Subtract 4 × (equation 2) from equation 1:
{3 x+0 y+2 z = -76 |     (equation 1)
0 x+y+0 z = 19 |     (equation 2)
0 x+0 y+z = -23 |     (equation 3)
Subtract 2 × (equation 3) from equation 1:
{3 x+0 y+0 z = -30 |     (equation 1)
0 x+y+0 z = 19 |     (equation 2)
0 x+0 y+z = -23 |     (equation 3)
Divide equation 1 by 3:
{x+0 y+0 z = -10 |     (equation 1)
0 x+y+0 z = 19 |     (equation 2)
0 x+0 y+z = -23 |     (equation 3)
Collect results:
Answer: | {x = -10   y = 19    z = -23

Do you intersted to explain to you how to solve it by matrices ?