Please help, Solve Systems of Linear Equations

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

Step-by-step explanation using elimination:

Solve the following system:

{3 x + 2 y - z = 8 | (equation 1)

2 z + 2 x = -4 | (equation 2)

3 y + x = 4 | (equation 3)

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

{3 x + 2 y - z = 8 | (equation 1)

0 x - (4 y)/3 + (8 z)/3 = -28/3 | (equation 2)

x + 3 y+0 z = 4 | (equation 3)

Multiply equation 2 by 3/4:

{3 x + 2 y - z = 8 | (equation 1)

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

x + 3 y+0 z = 4 | (equation 3)

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

{3 x + 2 y - z = 8 | (equation 1)

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

0 x+(7 y)/3 + z/3 = 4/3 | (equation 3)

Multiply equation 3 by 3:

{3 x + 2 y - z = 8 | (equation 1)

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

0 x+7 y + z = 4 | (equation 3)

Swap equation 2 with equation 3:

{3 x + 2 y - z = 8 | (equation 1)

0 x+7 y + z = 4 | (equation 2)

0 x - y + 2 z = -7 | (equation 3)

Add 1/7 × (equation 2) to equation 3:

{3 x + 2 y - z = 8 | (equation 1)

0 x+7 y + z = 4 | (equation 2)

0 x+0 y+(15 z)/7 = -45/7 | (equation 3)

Multiply equation 3 by 7/15:

{3 x + 2 y - z = 8 | (equation 1)

0 x+7 y + z = 4 | (equation 2)

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

Subtract equation 3 from equation 2:

{3 x + 2 y - z = 8 | (equation 1)

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

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

Divide equation 2 by 7:

{3 x + 2 y - z = 8 | (equation 1)

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

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

Subtract 2 × (equation 2) from equation 1:

{3 x + 0 y - z = 6 | (equation 1)

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

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

Add equation 3 to equation 1:

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

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

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

Divide equation 1 by 3:

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

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

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

Collect results:

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

answer: 48

just remember to use pemdas:

parentheses

expressions

multiplication

division

addition

subtraction

maybe i can