I need some help with the following, apparently simplistic, system. I was hoping someone here may be able to firgure it out and write out in what manner you find the solution set.
4x + 6y + 4z = 30
6x + y + z = 16
3x + 2y + 12z = 12

I need some help with the following, apparently simplistic, system. I was hoping someone here may be able to firgure it out and write out in what manner you find the solution set.
4x + 6y + 4z = 30
6x + y + z = 16
3x + 2y + 12z = 12
Do you know linear algebra?
I'm very familiar with solving systems of equations, but I can't for the life of me figure this one. An explanation would be much appreciated.
Set to 0 then Solve for x using the first equation, then substitute in for all the x's in the other 2 equations. Then repeat the same but solving for y and z respectively.
Its usually quicker using matrices. Are you familiar with matrix and vectors?
Yea, but I'm still stuck.
OK start by moving the 20, 16 and 12 over to the other side.......
I was hoping someone would just solve and then explain.
OK I will explain, but its going to take a lot explaining with these bastids.
Set to 0:
4x + 6y + 4z 30=0
6x + y + z  16=0
3x + 2y + 12z 12=0
....You with so far?
Yes, though I've never tried this approach.
OK its better to use the 2nd equation and solve for "y" 1st.
So:
6x+z+y16=0 ...bring the y over gives
6x+z16=y ......so divide out the 1 on the y
6/1x+z/116/1=y ..........simplified gives
6x2+16=Y
........Yeah?
...You still with?
This is a very alien method to me.
What method do you use then? I think this is called the substition method, but its been a while.
Evidently, the wrong method! :P
Look you moron, Im trying to help you. Dont do dishonest insults on me.
Can you even solve elementary equations?
It was just a joke, sheesh. But like I said, I'm very familiar with these types of equations, it's just that this one is giving me considerable trouble.
For the most part, I would just like someone to straight up solve it, and show the steps so I can move on, instead of arbitrarily going back and forth stacking up post counts. Dig?
I was being kind in not saying so, but you haven't a clue, have you?
I've solved for Y, now sub it in. That should be enough for someone who understands linear systems to procede.
I have better things to do than peoples homework, then get rubbished for the effort.
What ever retard!Originally Posted by Guaged
You're a touche one. If you can tell me the solution set (x, y, z), then I'll feel that there is some substance to your odd disaffection from the topic at hand.
Just as an aside, I'm almost ashamed to say, but I spent roughly 2 hours on this equation already so there is likely one of two things wrong; my technique, or simply the numbers I used to cancle out a given variable in some arbitrary step.
If you had any patience you would have had the answer plus a step by step walkthrough to copy down.
Go fish! smart ass! Post it on a homework doing for ya site.
It's not homework, per se, I just like to practice math so I'm not too rusty when I get back to school. Though, you're correct, I havn't a clue about how to solve this certain equation.
Yeah actually I think the canceling method is the method to use with 3 variables.
To be honest I've spent hours on these even when they were fresh on my mind, and Im tired.
Look I'm sorry I couldn't help.
Well I tried substituting and solved for x a couple different ways, got the result of x = 350/169 both times. (Haven't bothered to solve for y and z as that's pretty straightforward after solving for x) Is this a homework/textbook type problem that's supposed to get neat tidy answers?
Yea, I'm very sure the solution is just whole numbers.Originally Posted by Neutrino
I'm fairly certain I set up the system wrong from the beginning, after a little toying around with the signs I've figured it out, lol. (2, 3, 1)
That doesn't work for 3x + 2y + 12z = 12Originally Posted by Guaged
Presicely. The third equation needed to be:Originally Posted by Harold14370
3x  2y + 12z = 12,
I had to set it up from a very vauge word problem, so there in lies my mistake. I hate being confused by things so simple.
In one of my maths texts, there are some 'trick questions' where the answer is just simply, these equations can not exist at the same time. Although in saying that I'm not saying that's true for this case, because I can't be bothered working it out.
lol did u actually do it? or just plug and chug?
X = 2 . Y = 3 . Z = 1
Solved by multiplying through to get two equations with say 4Z in them. Then subtracting the equations to eliminate Z. Same process with a different pair of equations so you are down to just 2 equations in X and Y.
3X +2Y + 12Z = 12 gives funny numbers.. Works with 2Y.
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