how does 3+2(-11/14) become 3-11/7?

- anonymous

how does 3+2(-11/14) become 3-11/7?

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- anonymous

14 is cancelled by table 2 as 2*7=14

- xkehaulanix

According to PEMDAS, the order in which you do operations is such:
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction
Since you have multiplication in the problem, you would do the multiplication before the addition.
3 + 2 x (-11/14)
The 2 cancels part of the 14 because it is factorable by 2:
2 x -11/(2 x 7) = -11/7
Since the 3 is unaffected by the multiplication, it remains in the equation.

- anonymous

I think I understand.

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- anonymous

thats gud.

- anonymous

so how would you proceed to the next step? x=(3-11/7)-2z.......How do you solve 3-11/7

- anonymous

to solve 3-11/7
taking L.C.M,
=(21-11)/7
=11/7

- anonymous

sory its 10/7

- anonymous

l.c.m?

- anonymous

so does 3-11/7 become 3/1*11/7 and cross multiply 3and 7, 1 and 11?

- anonymous

but then wouldn't the answer be 3?

- anonymous

yes l.c.m

- anonymous

l.c.m is of the numbers in denominatot

- anonymous

3/1-11/7 so we have 1 and 7 in denominator so 7 is the l.c.m

- anonymous

no its not cross multiplication i just took l.c.m

- xkehaulanix

l.c.m stands for "Least Common Multiple," but you can call it the least common denominator too. To add 3 to 11/7, you need to make sure that the denominators are the same.
So yes, 3 is currently divided by 1, and 11 is divided by 7. You can't cross multiply for this part though because the numbers are not in a proportion.
So to make the denominator on 3 equal to the denominator under 11, you would need to multiply it by 7.
So 3/1 x 7/7= 21/7

- anonymous

I'm sorry, I don't even understand what LCM is, I haven't done math in 15 years

- anonymous

i got it what are u sayng,its ok in ur sense cross multiply after l.c.m

- anonymous

ok dnt worry i tell what the l.c.m is

- anonymous

so then it becomes 21/7-11/7?

- anonymous

and then 10/7 okay!

- xkehaulanix

Yup! That's it!

- anonymous

so now what is the next step, x=10/7-2z

- anonymous

as on present case we have 1,7 so
we divide both on a single number until we get 1 then multiply those numbers which is l.c.m

- xkehaulanix

Are you solving for z?

- anonymous

ok its 2z?solving 4 z or x?

- anonymous

ummmm, this was the original equation,
4x+6y-8z=1
-6x-9y+12z=0
x-2y-2z=3

- xkehaulanix

Ohh. You're supposed to solve for all of the variables, right?

- anonymous

ok u have system of linear equations to solve

- anonymous

yes, w/ help I got this far,
x=3-2y-2z
4(3-2y-2z)+6y-8z=1
-6(3-2y-2z)-9y+12z=0
and then
4(3-2y-2z)+6y-8z=1
12+8y+8z+6y-8z=1
12+14y=1
14y=-11
y=-11/14
so
x=3+2(-11/14)-2z
x=(3-11/7)-2z
and so now I'm at x=10/7-2z

- anonymous

so now I know y=10/7 or 1 3/7?

- xkehaulanix

Wow..that looks like a lot to type. Hmm, ok.
There's a small error in the beginning. The third equation you wrote was x-2y-2z=3, right? In the last post you made, you replaced x with 3-2y-2z. Those equations are actually not equal, but it's just a problem with positive and negative signs.
x-2y-2z=3
Since you want x separated from the other two variables, you would move them to the other side of the equation. First, you would add 2y to both sides to keep the equation balanced (remember, what you do to one side of an equation must be done to the other, just like a seesaw).
So add 2y and 2z to both sides to separate x.
x=3+2y+2z

- anonymous

okay, so then it becomes 4(3+2y+2z) +6y-8z=1
12+8y+8z+6y-8z=1
12+14y=1.....so it still comes out the same in the end doesn't it? Am I not still left with x=10/7-2z?

- anonymous

OOO no, it changes it to x=10/7+2z

- xkehaulanix

Whoa, you lost me xD
Well from there you can get y, which is still correct. And if you plug y into that equation, it becomes
x=3+2(-11/14)+2z
So yes, the last one you posted was right.

- anonymous

okay, so how do I solve it? how do I solve x=10/7+2z?

- anonymous

do I now know that y=10/7? how do I figure out x and z?

- xkehaulanix

Okay, so far you've only had to use two equations. You can plug in the new x value (10/7 + 2z) into the last equation to solve for z since you already know that y is equal to -11/14.
The y you got from the 12+14y=1 that you solved earlier. So you already know what y is!

- anonymous

so now the equation is -6(10/7+2z)-9(10/7)+12z=0?

- xkehaulanix

Err...yes to everything except the 10/7. The equation for y you had earlier was
12+14y=1
When you solved for it, you ended up with -11/14.

- anonymous

oh....ok so 10/7 only works in the x-2y-2z=3 equation?

- anonymous

or the 4x+6y-8z=1 equation?

- anonymous

I thought I was getting it, but now I'm confused all over again

- xkehaulanix

The 10/7 you got in that equation wasn't equal to any of the variables, actually. It was just part of the equation for x. The equation was x=10/7 + 2z, right? All the 10/7 tells you is that x is 10/7 larger than twice of z. It doesn't stand for a variable.
xD Math is such a joy, isn't it?

- anonymous

lol yea

- anonymous

okay. so y=-11/14

- anonymous

so now I do x=2(-11/7)+2z?

- anonymous

or 4(3+2(-11/7)+2z)+6(-11/14)-8z=1 and -6(3+2(-11/14)+2z)-9(-11/14)+12z=0?

- anonymous

becoming 12+8(-11/14)+6(-11/14)+4z=1and -18-12(-11/14)-9(-11/14)=0

- xkehaulanix

Whoops, sorry. I had to go find the equations again.
Whoaaah. You could do that, but trying to keep your variables straight gets really confusing, doesn't it?
When it comes to multiple equations like this, it's usually easier to try and separate a variable before putting in more values. So instead of plugging in the values for x and y in the beginning, try separating x or z from one of the equations first.
Say we took 4x+6y-8z=1
We could pull out the z to make it a little less confusing.
So 4x+6y=1+8z
4x+6y-1=8z
And you could divide both sides by 8 if you really wanted to, but it depends on how comfortable you are with fractions...and on a computer screen, fractions are hard to keep track of.
Then for y you would plug in -11/14

- anonymous

I'm good with fractions for the most part, it's the rest of this junk

- anonymous

and I'm keeping track on paper too :0)

- anonymous

so it would become 4x+6(-11/14) -1=8z?

- xkehaulanix

And for x, you could plug in either the original equation of
x=3+2y+2z or the simpler x=10/7+2z that you figured out earlier.
Oh, then you shouldn't have any problems xD I think I'm avoiding fractions for myself, just because I never see the division sign...

- xkehaulanix

Yup. It has a lot less numbers than before, right? xD

- anonymous

okay....so 4(10/7+2z) +6(-11/14)-1=8z

- anonymous

yea it does

- anonymous

okay, so I figured the first half, 40+8z, how do you multiple a positive with a negative?6(-11/14)

- xkehaulanix

To multiply a negative number, the rules are pretty simple. If you multiply a negative by a positive (as in this case), the result will be a negative number. It multiplies the same as two positives though. If you have two negative numbers, then the result will be positive instead.

- anonymous

aha, so then 6(-11/14) becomes -66

- anonymous

so then the equation becomes 2z-27=8z

- xkehaulanix

Yep, but don't forget to divide by the 14.
OH NO. I'm sorry, I wasn't paying attention to the equation I picked! We already used this equation to solve for the x value, so the variables are just going to cancel out...You have to use the third equation to solve for z.
So use -6x-9y+12z=0...
Ahh...my bad.

- anonymous

oh okay, hold on. let me try

- anonymous

okay, so:
-6x-9y+12z=0
-6x-9y=-12z
-6(10/7)+2z)-9(-11/14)=-12z
-60+2z+99=-12z
39+2z=-12z
39=-12z-2z
39=-14z?

- anonymous

so z=2 -11/14

- anonymous

or more like -2 11/14

- xkehaulanix

Ahh...almost. You have the equation right, but somewhere in the middle of multiplying you dropped your denominators.
Take 6[(10/7)+2z], for example.
When you multiply by the 6, you should end up with (60/7)+12z, rather than 60+12z.

- anonymous

ummmm....6(10/7) ooooohhh okay

- xkehaulanix

Mmhmm. The same goes for the -9(-11/14). You got the positive sign right on that part though :D

- anonymous

okay, i'll try again

- anonymous

so -6 is a whole number, so to multiply it by 10/7 you have to make it a fraction correct? so it becomes -6/1x10/7. then you have to have common denominators and what you do to the bottom you do to the top? -42/7x10/7 then becomes -420/7 becoming -60?

- anonymous

or was that other person wrong, and it's just -6/1x10/7=-60/7?

- anonymous

-8 4/7(simplified)

- xkehaulanix

The least common denominator is really only used when adding and subtracting. So the other person was right on that part. For multiplication, the last thing you posted is correct.
I'd be careful with mixed fractions (the whole number with the fraction) only because multiplying them gets tricky.
And...I think something went wrong again because my variable disappeared. Let me try doing the problem on paper ^-^'

- anonymous

okay, i'm working it out on paper too, and this is where i'm at,
-60/7+2z-99/14=-12z
-219/14+2z=-12z
-219/14=-12z-2z
-219/14=-14z

- anonymous

so now I divide -219/14 by -14 to =z?

- anonymous

-219/14 divided by-14/1=-219/14x-1/14=219/196

- xkehaulanix

There's one more thing missing in your multiplication. When you multiply something inside of parenthesis, you have to multiply everything inside.
So you had 6[(10/7)+2z], right? You have to multiply both the 10/7 AND the 2z by 6 because they're inside the parenthesis. It goes back to the PEMDAS thing I mentioned a long time ago.

- anonymous

crap. I thought I did that.

- anonymous

okay so -60/7-12z+99/14=-12z

- anonymous

therefore, -3/2-12z=-12z

- xkehaulanix

Yup, that's what I've been getting. But from there, the z keep getting dropped, no matter which equation I use..

- anonymous

-3/2=-12z+12z

- anonymous

yea, it did it for me too, so does that mean there is no solution? we find out x and y but never z?

- xkehaulanix

Technically, you can find z if you find x. But we shouldn't be dropping z in every equation we try because...well, it's not really logical, I guess. I'm still looking to see if I messed up too.

- anonymous

okay

- xkehaulanix

Ahh, no good. I'm running in circles now. Are you sure the original equations are right?

- anonymous

That's what I was given. :0( Well thank you for all your help, I think it's time to call it a night. I will have to double check the equation tomorrow, and try again if the originals were wrong. Thank you for your patience, at least I know how to figure it out next time!

- xkehaulanix

Okay then. Sorry I couldn't help more though!

- anonymous

It's okay, you were great! Thank you!

- xkehaulanix

No problem! Good luck tomorrow!

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