Solve the following system of equations for all three variables

Question

2x-3y-2z=7

8x-3y+2z=-9

4x+3y+4z=-3

Answer 1

Eq 1) 2x - 3y - 2z = 7

Eq 2) 8x - 3y + 2z = -9

Eq 3) 4x + 3y + 4z = -3

Notice the the y-variable for Eq 1 and 2 has a "-3" coefficient and the y-variable for Eq 3 has a "+3" coefficient? So we can eliminate the y-variable if we add Eq 1 and Eq 3 and then add Eq 2 and 3 as follows:

Eq 1) 2x - 3y - 2z = 7

Eq 3) 4x + 3y + 4z = -3

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Eq 4) 6x + 2z = 4

Eq 2) 8x - 3y + 2z = -9

Eq 3) 4x + 3y + 4z = -3

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Eq 5) 12x + 6z = -12

Now, using Eq 4 and Eq 5, we can eliminate the x-variable if we multiply both sides of Eq 4 by "-2":

Eq 4 (modified): -12x - 4z = -8

Adding to Eq 5 we get:

-12x - 4z = -8

12x + 6z = -12

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2z = -20

z = -10

Plugging that into Eq 5 we get:

Eq 5) 12x + 6z = -12

12x + 6(-10) = -12

12x - 60 = -12

12x = 48

x = 4

Plugging x = 4 and z = -19 into Eq 1 we get:

Eq 1) 2x - 3y - 2z = 7

2(4) - 3y -2(-10) = 7

8 - 3y + 20 = 7

-3y = -21

y = 7

So x = 4, y = 7, and z = -10