3 Variable System of Equations

Question

Answer 1

2x + y + z = 1 EQ1

2x - y - 3z = -3 EQ2

x -2y -4z = -2 EQ3

Subtract 2 time EQ3 from EQ1

5y + 9z = 5 EQ4

Subtract 2 times EQ3 from EQ2

3y + 5z = 1 EQ5

Solve for z in EQ5: z = (1-3y)/5

Substitute for z in EQ4

5y + 9(1-3y)/5 = 5

Solve for y

5y + (9 -27y)/5 = 5

25y + 9 -27y = 25

-2y = 25 -9 = 16

y = 16/-2 = -8

Answer 2

option 1: solves the system outright

eq1 - eq2

===========

2y + 4z = 4

y + 2z = 2 <--- divides by 2

-2 * eq3 + eq1

================

5y + 9z = 5

2 x 2 system

===================

y + 2z = 2

5y + 9z = 5

-5 * eq1 + eq2

====================

-z = -5

z = 5

substitutes: new 2x2

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2x + y = -4

2x - y = 12

x - 2y = 18

eq1 - eq2

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2y = -16

y = -8

substitutes:

2x + -8 + 5 = 1

2x - 3 = 1

2x = 4

x = 2

solution x=2, y = -8 z = 5

the solution checks, so y=-8

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option 2: plug an answer so as to find

which one produces a consistent solution

first answer: y=5

2x + z = -4

2x - 3z = 2

x - 4z = 8

subtracts eq1 - eq2:

4z = -6 --> z = -3/2

-2*eq3 + eq2:

5z = -14 ---> z = -14/5

contradiction: y = 5 eliminated

skips to last answer: y=-10

2x + z = 11

2x - 3z = 7

x - 4z = -22

subtracts eq1-eq2:

4z = -4 --> z = -2

-2*eq3+eq2:

5z = 51 --> z = 51/2

contradiction: y=-10 eliminated

option #2: y = -8

2x + z = 9

2x - 3z = -11

x - 4z = -18

subtracts eq1-eq2:

4z = 20 ---> z = 5

-2*eq3 + eq2:

5z = 25 --> z = 5

substitutes into 1st equation:

2x + -8 + 5 = 1

2x - 3 = 1

2x = 4

x = 2

plugs everything into 2nd equation:

2(2) - -8 - 3(5) =

4 + 8 - 15 = -3 yes it checks

plugs everthing into 3rd equation:

2 - 2(-8) - 4(5) =

2 - -16 - 20 =

18-20 = -2

so yes it check

the solution is (x=2, y=-8, z=5)