Pre-Calculus – Systems of Equations with 3 Variables please show work/steps x-3z=7 2x+y-2z=11 -x-2y+9z=13 Question Pre-Calculus – Systems of Equations with 3 Variables please show work/steps x-3z=7 2x+y-2z=11 -x-2y+9z=13 in progress 0 Math Adeline 3 weeks 2022-01-07T15:42:38+00:00 2022-01-07T15:42:38+00:00 1 Answer 0 views 0

## Answers ( )

Answer:x = 10, y = -7

, z = 1

Step-by-step explanation:Solve the following system:

{x – 3 z = 7 | (equation 1)

2 x + y – 2 z = 11 | (equation 2)

-x – 2 y + 9 z = 13 | (equation 3)

Swap equation 1 with equation 2:

{2 x + y – 2 z = 11 | (equation 1)

x + 0 y – 3 z = 7 | (equation 2)

-x – 2 y + 9 z = 13 | (equation 3)

Subtract 1/2 × (equation 1) from equation 2:

{2 x + y – 2 z = 11 | (equation 1)

0 x – y/2 – 2 z = 3/2 | (equation 2)

-x – 2 y + 9 z = 13 | (equation 3)

Multiply equation 2 by 2:

{2 x + y – 2 z = 11 | (equation 1)

0 x – y – 4 z = 3 | (equation 2)

-x – 2 y + 9 z = 13 | (equation 3)

Add 1/2 × (equation 1) to equation 3:

{2 x + y – 2 z = 11 | (equation 1)

0 x – y – 4 z = 3 | (equation 2)

0 x – (3 y)/2 + 8 z = 37/2 | (equation 3)

Multiply equation 3 by 2:

{2 x + y – 2 z = 11 | (equation 1)

0 x – y – 4 z = 3 | (equation 2)

0 x – 3 y + 16 z = 37 | (equation 3)

Swap equation 2 with equation 3:

{2 x + y – 2 z = 11 | (equation 1)

0 x – 3 y + 16 z = 37 | (equation 2)

0 x – y – 4 z = 3 | (equation 3)

Subtract 1/3 × (equation 2) from equation 3:

{2 x + y – 2 z = 11 | (equation 1)

0 x – 3 y + 16 z = 37 | (equation 2)

0 x+0 y – (28 z)/3 = (-28)/3 | (equation 3)

Multiply equation 3 by -3/28:

{2 x + y – 2 z = 11 | (equation 1)

0 x – 3 y + 16 z = 37 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 16 × (equation 3) from equation 2:

{2 x + y – 2 z = 11 | (equation 1)

0 x – 3 y+0 z = 21 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 2 by -3:

{2 x + y – 2 z = 11 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{2 x + 0 y – 2 z = 18 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Add 2 × (equation 3) to equation 1:

{2 x+0 y+0 z = 20 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by 2:

{x+0 y+0 z = 10 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer: {x = 10

, y = -7

, z = 1