Question

3×3 Systems Elimination by Addition

1) 4x-2y-2z=2
    -x+3y+2z=-8
    4x-5y+z=11

2)-5x-2y+20z=-28
      2x-5y+15z=-27
     -2x-2y-5z=-12

Please show every step in clear handwriting, so I can figure out how to do it myself.

Answer 1

1)

4x-2y-2z=2
    -x+3y+2z=-8
    4x-5y+z=11

Multiply second eqaution by 4 and add with first

4x-2y-2z=2
-4x +12y+8z = -32

so,

10y +6z =-30

5y+3z=-15

Subtract first and third equation to obtain

3y-3z= -9

y-z = -3

5y-5z = -15

So, equations are

5y+3z=-15

5y-5z = -15

Subtract both

8z =0

z=0

substitute z= 0 in 5y+3z=-15

5y= -15

y= -3

Substitute y= -3 and z =0 in   -x+3y+2z=-8

-x -9 +0 = -8

-x = 1

x= -1

Hence, solutions are

x = -1, y = -3, z = 0

Please open new post for second question.