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14 is cancelled by table 2 as 2*7=14
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.
I think I understand.
so how would you proceed to the next step? x=(3-11/7)-2z.......How do you solve 3-11/7
to solve 3-11/7 taking L.C.M, =(21-11)/7 =11/7
sory its 10/7
so does 3-11/7 become 3/1*11/7 and cross multiply 3and 7, 1 and 11?
but then wouldn't the answer be 3?
l.c.m is of the numbers in denominatot
3/1-11/7 so we have 1 and 7 in denominator so 7 is the l.c.m
no its not cross multiplication i just took l.c.m
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
I'm sorry, I don't even understand what LCM is, I haven't done math in 15 years
i got it what are u sayng,its ok in ur sense cross multiply after l.c.m
ok dnt worry i tell what the l.c.m is
so then it becomes 21/7-11/7?
and then 10/7 okay!
Yup! That's it!
so now what is the next step, x=10/7-2z
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
Are you solving for z?
ok its 2z?solving 4 z or x?
ummmm, this was the original equation, 4x+6y-8z=1 -6x-9y+12z=0 x-2y-2z=3
Ohh. You're supposed to solve for all of the variables, right?
ok u have system of linear equations to solve
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
so now I know y=10/7 or 1 3/7?
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
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?
OOO no, it changes it to x=10/7+2z
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.
okay, so how do I solve it? how do I solve x=10/7+2z?
do I now know that y=10/7? how do I figure out x and z?
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!
so now the equation is -6(10/7+2z)-9(10/7)+12z=0?
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.
oh....ok so 10/7 only works in the x-2y-2z=3 equation?
or the 4x+6y-8z=1 equation?
I thought I was getting it, but now I'm confused all over again
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?
okay. so y=-11/14
so now I do x=2(-11/7)+2z?
or 4(3+2(-11/7)+2z)+6(-11/14)-8z=1 and -6(3+2(-11/14)+2z)-9(-11/14)+12z=0?
becoming 12+8(-11/14)+6(-11/14)+4z=1and -18-12(-11/14)-9(-11/14)=0
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
I'm good with fractions for the most part, it's the rest of this junk
and I'm keeping track on paper too :0)
so it would become 4x+6(-11/14) -1=8z?
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...
Yup. It has a lot less numbers than before, right? xD
okay....so 4(10/7+2z) +6(-11/14)-1=8z
yea it does
okay, so I figured the first half, 40+8z, how do you multiple a positive with a negative?6(-11/14)
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.
aha, so then 6(-11/14) becomes -66
so then the equation becomes 2z-27=8z
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.
oh okay, hold on. let me try
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?
so z=2 -11/14
or more like -2 11/14
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.
ummmm....6(10/7) ooooohhh okay
Mmhmm. The same goes for the -9(-11/14). You got the positive sign right on that part though :D
okay, i'll try again
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?
or was that other person wrong, and it's just -6/1x10/7=-60/7?
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 ^-^'
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
so now I divide -219/14 by -14 to =z?
-219/14 divided by-14/1=-219/14x-1/14=219/196
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.
crap. I thought I did that.
okay so -60/7-12z+99/14=-12z
Yup, that's what I've been getting. But from there, the z keep getting dropped, no matter which equation I use..
yea, it did it for me too, so does that mean there is no solution? we find out x and y but never z?
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.
Ahh, no good. I'm running in circles now. Are you sure the original equations are right?
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!
Okay then. Sorry I couldn't help more though!
It's okay, you were great! Thank you!
No problem! Good luck tomorrow!