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sashaice [31]
3 years ago
11

X+y+z=12 6x-2y+z=16 3x+4y+2z=28 What does x, y, and z equal?

Mathematics
1 answer:
lianna [129]3 years ago
8 0

Answer:

x = 20/13 , y = 16/13 , z = 120/13

Step-by-step explanation:

Solve the following system:

{x + y + z = 12 | (equation 1)

6 x - 2 y + z = 16 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Swap equation 1 with equation 2:

{6 x - 2 y + z = 16 | (equation 1)

x + y + z = 12 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Subtract 1/6 × (equation 1) from equation 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+(4 y)/3 + (5 z)/6 = 28/3 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Multiply equation 2 by 6:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

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

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

0 x+5 y + (3 z)/2 = 20 | (equation 3)

Multiply equation 3 by 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

0 x+10 y + 3 z = 40 | (equation 3)

Swap equation 2 with equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+8 y + 5 z = 56 | (equation 3)

Subtract 4/5 × (equation 2) from equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+(13 z)/5 = 24 | (equation 3)

Multiply equation 3 by 5:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+13 z = 120 | (equation 3)

Divide equation 3 by 13:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Subtract 3 × (equation 3) from equation 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y+0 z = 160/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Divide equation 2 by 10:

{6 x - 2 y + z = 16 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Add 2 × (equation 2) to equation 1:

{6 x + 0 y+z = 240/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Subtract equation 3 from equation 1:

{6 x+0 y+0 z = 120/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Divide equation 1 by 6:

{x+0 y+0 z = 20/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Collect results:

Answer:  {x = 20/13 , y = 16/13 , z = 120/13

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