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Show work. Solve using the substitution method

1. 5x + 12y + z = 10
2x + 5y + 2z = -1
x + 2y - 3z = 5

2. 2x - 4y + z =3
X - 3y + z= 5
3x-7y+2z=12

3. 5x + 8y - 6z=14
3x+ 4y -2z=8
x+2y-2z=3

4. 5x-11y+6z=12
-x+3y-2z=-4
3x-5y+2z=4

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

6. 2x-y-z=0
x+2y+z=3
3x+4y+2z=8

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Solution Summary

The substitution method is applied.

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Problem #1
5x + 12y + z = 10 (1)
2x + 5y + 2z = -1 (2)
x + 2y - 3z = 5 (3)
From (1), we get z = 10 - 5x - 12y, then we substitute z in (2) and (3), then get
2x + 5y + 2(10 - 5x - 12y) = -1 => 8x + 19y = 21 (4)
x + 2y - 3(10 - 5x - 12y) = 5 => 16x + 38y = 35 (5)
We note (4)x2: 16x + 38y = 42
Compared with (5), we get a contradiction.
So the system is inconsistent and thus has no solution.

Problem #2
2x - 4y + z =3 (1)
X - 3y + z = 5 (2)
3x-7y + 2z=12 (3)
From (1), we get z = 3 - 2x + 4y, then we substitute z in (2) and (3), then get
x - 3y + 3 - 2x + 4y = 5 => -x + y = 2 (4)
3x - 7y + 2(3 - 2x + 4y) = 12 ...

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