Question

solve the system of equations. I genuinely have no clue how to do this so if someone could help explain it, that would be fantastic!
1) x+y-z=-1
x+y+z=3
3x-2y-z=4
2)x+y=5
3x+z=2
4y-z=8
3) 2x-y+z=-1
x-y+z=1
x-2y+z=2
4) x-z=5
y+3z=12
2x+y=7

Answer 1

Well these are actually really fun. Let's look at number 1.

What equation there seems like the simplest to solve for a variable? (Which one will require the least amount of step/or is easiest for you) to solve for one of the variables in it.

Answer 2

Ill help with number 2 since @Cosmichaotic will probably help you with number 1.

Answer 3

Haha, nice.

Answer 4

we can both help :D

Answer 5

Of course, I prefer that. I make a lot of mistakes and need supervision.

Answer 6

ok, for number 2. you want to end up finding what values for x, y, and z make each equation true. so what you probably want to do is focus first on solving for z in the second equation

Answer 7

@cosmichaotic Well the last one has more numbers in it so it seems fairly straightforward I guess

Answer 8

to do that, subtract your 3x from both sides so you have z = -3x +2

Answer 9

next, look at your equation 3 for number 2 and plug in (-3x+2) for z

Answer 10

Alright so it's your choice, do you want to solve for x, y or z?

Answer 11

More numbers means more operations, let's choose the first equation since it seems simple enough to isolate one of the variables.

Answer 12

what you should be looking at now is something such as this \[4y - (-3x + 2) = 8\]

Answer 13

with all your negatives multiplied out you should end up with this: \[4y + 3x - 2 = 8\]

Answer 14

@cosmichaotic okay, that makes sense... So we just start to solve for x?

Answer 15

@technosoul And then I just solve for the next variable and plug it in to the next equation?

Answer 16

pretty much, yes

Answer 17

@technosoul Thank you so much! You have no idea how much it helped

Answer 18

sure thing. just so you know though, number two is going to have a slightly confusing answer and i can explain that after you find your answer

Answer 19

ill start on number 3 though, for now.

Answer 20

Absolutely, sure, solve for x.
We will get:
x+y-z=-1 -> x = -1 -y + z

So now that we have x, (x = -1 -y + z), we can plug that into the x in the next equation.
Plugging in what we got for x: (x)+y+z=3 -> -1 -y + z + y + z = 3

Answer 21

Now we just have two variables to worry about. First, let's simplify it by combining like terms.

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

Adding the y's and z's,
where -y + y = 0 (We won't have to worry about y's)
and z + z = 2z

We have,
-1 +0y + 2z = 3 or -1 + 2z = 3

Adding 1 to both sides to combine the constants,

-1 + 2z = 3 -> 2z = 4

And we simplify this by dividing by 2 on both sides,

2z = 4 -> z = 2

Answer 22

I know that's a long explanation, but read through it and let me know if you have any questions about how we came to:

x = -1 -y + z
z = 2

So far.

Answer 23

alright, number 3. so start by solving for a variable (i hope its ok i chose to solve for z first). Your best option to solving for z is your first and your last equations and ill explain why. The first equation has a value of z that will cancel out the second equation's value of y. and therefore solving for your x. The last equation has a value of z that will cancel out the second problems value of x and therefore solving for your y.

Answer 24

i promise it will make more sense when i write it out.

Answer 25

so first solve for z for your first equation.
\[2x - y +z=-1\]
then you add y to both sides to get
\[2x + z = y-1\]
after that subtract the 2x from both sides and you'll get
\[z = -2x +y-1\]
make sense so far?

Answer 26

I think that to solve it from there, your second equation is the easiest to use so plug in what you have for z in, in the second equation. You should get:
\[x-y+(-2x+y-1)=1\]

Answer 27

as i mentioned earlier, this will now solve for your x because your y's will now cancel out

Answer 28

once you get that, tell me, i can check to make sure you got what i did