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ehidna [41]
4 years ago
6

Solve the following System of Equations 5x+4y-z=1 2x-2y+z=1 -x-y+z=2

Mathematics
1 answer:
podryga [215]4 years ago
3 0

Answer:

x = 0 , y = 1 , z = 3

Step-by-step explanation:

Solve the following system:

{5 x + 4 y - z = 1 | (equation 1)

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

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

Subtract 2/5 × (equation 1) from equation 2:

{5 x + 4 y - z = 1 | (equation 1)

0 x - (18 y)/5 + (7 z)/5 = 3/5 | (equation 2)

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

Multiply equation 2 by 5:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

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

Add 1/5 × (equation 1) to equation 3:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

0 x - y/5 + (4 z)/5 = 11/5 | (equation 3)

Multiply equation 3 by 5:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

0 x - y + 4 z = 11 | (equation 3)

Subtract 1/18 × (equation 2) from equation 3:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

0 x+0 y+(65 z)/18 = 65/6 | (equation 3)

Multiply equation 3 by 18/65:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

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

Subtract 7 × (equation 3) from equation 2:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y+0 z = -18 | (equation 2)

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

Divide equation 2 by -18:

{5 x + 4 y - z = 1 | (equation 1)

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

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

Subtract 4 × (equation 2) from equation 1:

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

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

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

Add equation 3 to equation 1:

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

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

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

Divide equation 1 by 5:

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

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

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

Collect results:

Answer: {x = 0 , y = 1 , z = 3

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