Solve the following 3 × 3 system. enter the coordinates of the solution below.
2x – 3y – 2z = 4
x + 3y + 2z = –7
–4x – 4y – 2z = 10

The system is:

i)    2x – 3y – 2z = 4
ii)    x + 3y + 2z = –7
iii)   –4x – 4y – 2z = 10 

the last equation can be simplified, by dividing by -2, 

thus we have:

i)    2x – 3y – 2z = 4
ii)    x + 3y + 2z = –7
iii)   2x +2y +z = -5 

The procedure to solve the system is as follows:

first use any pairs of 2 equations (for example i and ii, i and iii) and equalize them by using one of the variables:

i)    2x – 3y – 2z = 4   
iii)   2x +2y +z = -5 

2x can be written as 3y+2z+4 from the first equation, and -2y-z-5 from the third equation.

Equalize:  

3y+2z+4=-2y-z-5, group common terms:
5y+3z=-9   

similarly, using i and ii, eliminate x:

i)    2x – 3y – 2z = 4
ii)    x + 3y + 2z = –7

multiply the second equation by 2:

i)    2x – 3y – 2z = 4
ii)    2x + 6y + 4z = –14

thus 2x=3y+2z+4 from i and 2x=-6y-4z-14 from ii:

3y+2z+4=-6y-4z-14
9y+6z=-18

So we get 2 equations with variables y and z:

a)   5y+3z=-9 
b)   9y+6z=-18

now the aim of the method is clear: We eliminate one of the variables, creating a system of 2 linear equations with 2 variables, which we can solve by any of the standard methods.

Let's use elimination method, multiply the equation a by -2:

a)   -10y-6z=18 
b)   9y+6z=-18
   add the equations:

-10y+9y-6z+6z=18-18
-y=0
y=0,

thus :
9y+6z=-18 
0+6z=-18
z=-3

Finally to find x, use any of the equations i, ii or iii:

2x – 3y – 2z = 4 

2x – 3*0 – 2(-3) = 4

2x+6=4

2x=-2

x=-1

Solution: (x, y, z) = (-1, 0, -3 ) 

Remark: it is always a good attitude to check the answer, because often calculations mistakes can be made:

check by substituting x=-1, y=0, z=-3 in each of the 3 equations and see that for these numbers the equalities hold.