what is the solution of the system of equations?

2x+2y+3z=-6
3x+5y+4z=3
2x+3y+4z=-10

2x+2y+3z=-6 times 3

3x+5y+4z=3 times 2

6 x + 6 y + 9 z = -18
6 x + 10y + 8 z = 6
-----------------------subtract
0 x - 4 y + 1 z = -24

3x+5y+4z=3 times 2
2x+3y+4z=-10 times 3

6 x + 10y + 8 z = 6
6 x + 9 y + 12z = -30
------------------------subtract
0 x + 1 y - 4 z = 36
so

- 4 y + 1 z = -24 times 4
+ 1 y - 4 z = 36

-16y + 4 z = - 96
+ 1y - 4 z = 36
------------------ add
-15 y = -60
y = 4
now go back and get z and x

solve it using matrix inversion method.

AX=B, where A is the coefficient matrix
2 2 3
A= 3 5 4
2 3 4
X=[x y z]' and B =[-6 3 -10]'
then X=(A^-1)B,that is A inverse*B
answer is x=5, y=4, z=-8

To find the solution to this system of equations, we can use a method called Gaussian elimination or row reduction. Here's how you can solve it:

Step 1: Write down the augmented matrix of the system of equations.
The augmented matrix is obtained by writing down the coefficients of the variables and the constant terms on the right-hand side. For the given system of equations, the augmented matrix is:

[ 2 2 3 | -6 ]
[ 3 5 4 | 3 ]
[ 2 3 4 | -10 ]

Step 2: Perform row operations to transform the matrix into row-echelon form.
The goal is to simplify the coefficient matrix such that it is in a triangular form with leading 1's. Here's a series of row operations you can perform on the augmented matrix:

- Row2 = Row2 - (3/2) * Row1
- Row3 = Row3 - Row1

After performing these row operations, the augmented matrix becomes:

[ 2 2 3 | -6 ]
[ 0 1 1 | 12 ]
[ 0 -1 -2 | 4 ]

Step 3: Continue row operations to further simplify the matrix.
To eliminate the negative coefficient in the bottom left corner, we can perform the following row operation:

- Row3 = Row3 + Row2

After performing this row operation, the augmented matrix becomes:

[ 2 2 3 | -6 ]
[ 0 1 1 | 12 ]
[ 0 0 -1 | 16 ]

Step 4: Solve for the variables.
The augmented matrix is now in row-echelon form. From the last row of the matrix, we can deduce that -z = 16, which means z = -16.

Substituting this value of z back into the second row of the matrix, we get:
y + z = 12
y + (-16) = 12
y - 16 = 12
y = 12 + 16
y = 28

Substituting the values of y and z into the first row of the matrix, we get:
2x + 2y + 3z = -6
2x + 2(28) + 3(-16) = -6
2x + 56 - 48 = -6
2x + 8 = -6
2x = -6 - 8
2x = -14
x = -14 / 2
x = -7

Therefore, the solution to the system of equations is x = -7, y = 28, and z = -16.