Results 1 to 10 of 10

Math Help - I hope I am not bugging you all to much

  1. #1
    Newbie
    Joined
    Dec 2007
    Posts
    9

    I hope I am not bugging you all to much

    I have another problem that I can't for the life of me solve. I really hate math and doing math online is the absolute pits. I have never been so stressed in all my life.

    Thanks again for any help you can give.

    The augmented matrix of a system of three equations in the three unknowns x, y, and z is reduced to

    [2 1 1 | 1]
    [0 3 2 | 4]
    [0 0 0 | 0]

    Find the solution of the system.

    I was thinking z = 0 but the answer in the book is z = k k any integer.

    The answers are
    x = k + 1/6
    y = 4 - 2k/3
    z = k k any integer

    HOW????

    Did I mention I cannot stand math specially online math????
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
    Change that attitude of your's about math and you will do better in it, and also understand more of it. You do not understand because you are not interested in understanding it.
    Last edited by janvdl; December 15th 2007 at 09:31 PM. Reason: Grammar err
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2007
    Posts
    9

    That is so not true

    I want to understand it because I want to get a good grade in the class. But to understand it I have to be taught it and I cannot be taught it in and online class.

    Thank you though for taking your time to write to me your words of wisdom.
    Now how do I get the answer? That is what would really help not such so much the words of wisdom. Thank you
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,843
    Thanks
    320
    Awards
    1
    Quote Originally Posted by lakillian View Post
    I have another problem that I can't for the life of me solve. I really hate math and doing math online is the absolute pits. I have never been so stressed in all my life.

    Thanks again for any help you can give.

    The augmented matrix of a system of three equations in the three unknowns x, y, and z is reduced to

    [2 1 1 | 1]
    [0 3 2 | 4]
    [0 0 0 | 0]

    Find the solution of the system.

    I was thinking z = 0 but the answer in the book is z = k k any integer.

    The answers are
    x = k + 1/6
    y = 4 - 2k/3
    z = k k any integer

    HOW????

    Did I mention I cannot stand math specially online math????
    Your augmented matrix is telling you that you have two equations in three unknowns. That means you have one degree of freedom in the system.

    Your book is telling you that z can be any number. (It should be "any real number" not an integer.) Once you set z = k, then you can get values for x and y in terms of k. But for the record we can also construct solutions where x is any real number (then we have expressions for y and z in terms of the value of x), or y is any real number. These solutions will give equivalent answers, though in different forms.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2007
    Posts
    9

    Dan

    Thank you for the help I appreciate it but how did they come up with fractions as answers? That is what I don't understand sigh.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,843
    Thanks
    320
    Awards
    1
    Quote Originally Posted by lakillian View Post
    Thank you for the help I appreciate it but how did they come up with fractions as answers? That is what I don't understand sigh.
    From your augmented matrix we have the equations
    2x + y + z = 1
    and
    3y + 2z = 4

    I will simply let z = k where k is some real number. Thus
    2x + y + k = 1
    and
    3y + 2k = 4

    Solving the bottom equation for y I get
    y = \frac{4 - 2k}{3}

    and inserting this value of y into the top equation I get
    2x + \frac{4 - 2k}{3} + k = 1

    I get
    x = -\frac{k + 1}{6}

    Apparently I don't agree with the value of x you gave in your previous post. (Unless there was a typo and you left the negative sign off.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2007
    Posts
    98

    Thumbs up

    Quote Originally Posted by lakillian View Post
    I want to understand it because I want to get a good grade in the class. But to understand it I have to be taught it and I cannot be taught it in and online class.

    Thank you though for taking your time to write to me your words of wisdom.
    Now how do I get the answer? That is what would really help not such so much the words of wisdom. Thank you
    Now before you reply, finish reading it and please be humble about your attitude towards yourself.

    I guess we wear the same shoe, different size. You are right about understanding math and getting a good grade. That is the main reason I am here, but beyond the good grade there should be a point in which you are doing it for you and no the damned grade. I have a final on Monday and believe my chances at passing are very slim if none. This is my third time taking this course, and after countless battles I decided that I am passing this course one way or another.
    Even if it's in here, so far after looking around for $online math tutor$ I found this place and it is helping me more than the college math tutors and the ones I had to look for in the wanted ads.
    So far I was able to get a passing if not decent grade on my last test, to which is a good sign for my final....(hoping). I am able to get the help that I need here not from "professionals" but from people even fellow students.
    I don't regret my "math past" but I cant let that be my weakness, so I learn to not only for a good grade but to know as well, so I can pass down the same wisdom to those who might wear my shoes someday.

    Good luck on your next test/final! Don't let the grade get you down!!!!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Dec 2007
    Posts
    9

    Morzilla

    Thank you for taking the time to try to make me feel better. I really do appreciate it. I would love to understand this and be able to help others with it and even my daughter who is 12 right now and in Pre-Algebra. But she has a teacher in front of her. I do not have a teacher and the online one I have is more worried about real world examples than answering the questions that get posted about the tests and homework.

    That in itself is frustrating. And on top of that, I have two sons. 2 and 4 years and they constantly fight, bicker and call for me. I really don't get any help everyone else in the family is oblivious and ignores the children and I can't. My stress level has raised about 80% since I started this online class and i know I have gained about 8 lbs.

    I am trying really really hard to UNDERSTAND but it is really hard when you are teaching yourself with multiple distractions on top of that.

    Again Thank you very much and I will pray that you pass this class this time!!!!!

    God bless!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Dec 2007
    Posts
    9

    Dan I understand!!!!!!

    THank you so very much Now that I see it I can see how the answers were determined. I wish I could do this with every problem. You are really smart

    Thank you so much again

    Lakillian
    Quote Originally Posted by topsquark View Post
    From your augmented matrix we have the equations
    2x + y + z = 1
    and
    3y + 2z = 4

    I will simply let z = k where k is some real number. Thus
    2x + y + k = 1
    and
    3y + 2k = 4

    Solving the bottom equation for y I get
    y = \frac{4 - 2k}{3}

    and inserting this value of y into the top equation I get
    2x + \frac{4 - 2k}{3} + k = 1

    I get
    x = -\frac{k + 1}{6}

    Apparently I don't agree with the value of x you gave in your previous post. (Unless there was a typo and you left the negative sign off.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Dec 2007
    Posts
    98
    Quote Originally Posted by lakillian View Post
    Thank you for taking the time to try to make me feel better. I really do appreciate it. I would love to understand this and be able to help others with it and even my daughter who is 12 right now and in Pre-Algebra. But she has a teacher in front of her. I do not have a teacher and the online one I have is more worried about real world examples than answering the questions that get posted about the tests and homework.

    That in itself is frustrating. And on top of that, I have two sons. 2 and 4 years and they constantly fight, bicker and call for me. I really don't get any help everyone else in the family is oblivious and ignores the children and I can't. My stress level has raised about 80% since I started this online class and i know I have gained about 8 lbs.

    I am trying really really hard to UNDERSTAND but it is really hard when you are teaching yourself with multiple distractions on top of that.

    Again Thank you very much and I will pray that you pass this class this time!!!!!

    God bless!
    Thanks for the confidence! I wish you the same and hope that in time I will be able to help you pass the class as well!

    Kids at such age always want attention, and for some cases it's normal, but they should be able to understand that you need to do this.

    Don't worry you'll do good!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Hope someone can help
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: November 12th 2009, 06:27 AM
  2. This question has been bugging me for days!
    Posted in the Algebra Forum
    Replies: 3
    Last Post: June 11th 2009, 12:31 PM
  3. Replies: 2
    Last Post: October 20th 2008, 02:49 PM
  4. Simple Factoring question that is bugging me
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: June 12th 2008, 04:58 PM
  5. OMG .. I hope someone can do this.
    Posted in the Statistics Forum
    Replies: 6
    Last Post: March 5th 2008, 07:51 AM

Search Tags


/mathhelpforum @mathhelpforum