Systems of equations

Question

What is the solution for z in the following system of equations?

5x+2y-3z=17

2x+2y+z=5

3x+2y=-6

Answer 1

1) 5x + 2y - 3z = 17

2) 2x + 2y + z = 5

3) 3x + 2y = -6

Use elimination.

First, decide what variable to eliminate first

Since there is no "z" in equation 3), let's eliminate "z" from equations 1) and 2)

Multiply 2) by 3 to get:

2A) 6x + 6y + 3z= 15

Add to 1):

1) 5x + 2y - 3z = 17

2A) 6x + 6y + 3z= 15

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4) 11x + 8y = 32

Using equations 4) and 3) we can now eliminate the "y" by multiplying 3) by -4:

3A) -12x - 8y = 24

And adding to 4):

3A) -12x - 8y = 24

4) 11x + 8y = 32

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5) -x = 56

solving, we get x = -56

Using 3) and plugging in x = -56, we get:

3(-56) + 2y = -6

-168 + 2y = -6

2y = 162

y = 81

Using 2) and plugging in x = -56 and y = 81 we get:

2(-56) + 2(81) + z = 5

-112 + 162 + z = 5

50 + z = 5

z = -45

Answer 2

using the third equation we get y = -3+3x/2. Plugging that in to the second we get 2x - 6 +3x + z = 5, and so z = 11 - 5x. Now plug these in to the first equation and youll have only 1 variable to solve for, you can keep going in this manner until all variables are solved for.

Answer 3

I was a little slow to put the negative sign on at the end, but just wait like fifteen seconds and it shows up!