Easy shmeezy! This is a sort of lengthy answer, but hopefully I will really help you explain step by step what’s happening!

Ok, this is just like solving the two variable equations, except you obviously have three variables (x,y,z). The purpose of this is to solve for one variable at a time; that can only be accomplished when two of the variables cancel each other…now notice the first two equations:

2x – 3y + z = -1

+ x + 3y – z = 4

———————-

3x + 0 + 0 = 3 (-3y + 3y = 0; z + -z = 0)

3x = 3

x = 3/3

x = 1

Yay! You just solved for one of the variables! Now just plug it in and solve as you would a two variable equation:

x + 3y – z = 4

x – 2y + 2z = 3

Hmm…right now neither y nor z are cancelling…but wait! If I multiply the above equation by 2, I can get -2z and that will cancel +2z! Let’s see:

2(x + 3y – z = 4) Remember, you have to multiply everything!

2x + 6y – 2z = 8

x – 2y + 2z = 3

Now let’s substitute x = 1:

2(1) + 6y – 2z = 8

1 – 2y + 2z = 3

Now I can combine like terms (in this case, it’s just whole numbers!):

6y – 2z = 8 – 2 = 6

+ -2y + 2z = 3 -1 = 2

———————

4y + 0 = 8

4y = 8

y = 8/4

y = 2

Two down, one more to go! Since you now know two of the three variables, you can substitute in any of the three equations to get the value of z:

x + 3y – z = 4

1 + 3(2) – z = 4

1 + 6 – z = 4

7 – z = 4

-z = 4 – 7

-z = -3

z = -3/-1

z = 3

And there you have it! x = 1, y = 2, z = 3!

Hope this helps 🙂

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x=1

y=2

z=3

Hope that helps ya.

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