Page 1 of 3 123 LastLast
Results 1 to 15 of 33

Math Help - system of linear equations

  1. #1
    Junior Member
    Joined
    Aug 2006
    Posts
    38

    system of linear equations

    Hi guys after the lessons i have problem understanding and solving system of linear equations. I need help badly to understand how it actually work outs???

    pls help me

    solve the given systems of equations by the method of elimination by substitution

    33x + 2y = 34
    40y = 9x + 11

    by the method of elimination by addition or subtraction

    2x - 3y - 4 = 0
    3x + 2 = 2y
    Follow Math Help Forum on Facebook and Google+

  2. #2
    dan
    dan is offline
    Member dan's Avatar
    Joined
    Jun 2006
    From
    nebraska usa
    Posts
    113
    Quote Originally Posted by danielwu

    solve the given systems of equations by the method of elimination by substitution

    33x + 2y = 34
    40y = 9x + 11
    ok sove for x in terms of y in #1:
     x=\frac{34-2y}{33}
    then substitute that value in for x in #2
     40y=9(\frac{34-2y}{33})+11
    does that help???
    dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by danielwu

    by the method of elimination by addition or subtraction

    2x - 3y - 4 = 0
    3x + 2 = 2y
    You have,
    2x-3y=4 (1)
    3x-2y=-2 (2)
    Multiply (1) by 2 and (2) by 3 to get,
    4x-6y=8 (3)
    9x-6y=-6 (4)
    From (4) subtract (3),
    5x=-14
    Thus,
    x=-2.8
    From here you can solve for "y".
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Aug 2006
    Posts
    38

    Smile

    Quote Originally Posted by dan
    ok sove for x in terms of y in #1:
     x=\frac{34-2y}{33}
    then substitute that value in for x in #2
     40y=9(\frac{34-2y}{33})+11
    does that help???
    dan
    Hi i do know how to do the steps but i don't know how to slove it!

    how to slove this?-----> <br />
9(\frac{34-2y}{33})<br />
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2006
    Posts
    38
    Quote Originally Posted by ThePerfectHacker
    You have,
    2x-3y=4 (1)
    3x-2y=-2 (2)
    Multiply (1) by 2 and (2) by 3 to get,
    4x-6y=8 (3)
    9x-6y=-6 (4)
    From (4) subtract (3),
    5x=-14
    Thus,
    x=-2.8
    From here you can solve for "y".
    why do i have to multiply by 2 in (1)?
    &
    why do i have to multiply by 3 in (2)?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by danielwu
    why do i have to multiply by 2 in (1)?
    &
    why do i have to multiply by 3 in (2)?
    Because you are going to subtract the two resulting equations and eliminate
    the term in y. So you need to make the coefficient of y in each equation the
    same.

    In the first equation the coefficient of y is -3, and in the second is -2.
    So multiplying these equations by 2 and 3 makes the coefficient of y -6
    in both equations, so when the equations are subtracted the y terms is
    removed.

    RonL
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by danielwu
    Hi i do know how to do the steps but i don't know how to slove it!

    how to slove this?-----> <br />
9(\frac{34-2y}{33})<br />
    divide by nine and then multiply bt 33
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Aug 2006
    Posts
    38
    Quote Originally Posted by Quick
    divide by nine and then multiply bt 33
    my ghosh i still don't get it

    what divide by nine? then multiply by 33?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Aug 2006
    Posts
    38
    hi i am still quite confuse.
    it says solve the given systems of equations by either method of this section?

    30p = 55 - Q
    19p + 14Q + 3 = 0

    so in becomes
    30p + Q = 55
    19p + 14Q = -3
    and from equation (1) do i multiply by 14? to make the coefficient of Q in (1) the same as equation 2?

    420p + 14Q = 770
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by danielwu
    hi i am still quite confuse.
    it says solve the given systems of equations by either method of this section?

    30p = 55 - Q
    19p + 14Q + 3 = 0

    so in becomes
    30p + Q = 55
    19p + 14Q = -3
    and from equation (1) do i multiply by 14? to make the coefficient of Q in (1) the same as equation 2?

    420p + 14Q = 770
    That is correct, now subtract the second equation fron this to get:

    420p - 19p + 14Q - 14Q = 770 + 3,

    simplifying:

    401p = 773,

    p = 773/401.

    Now going back to your first equation:

    30p + Q = 55,

    so:

    Q=55-30p=55-30.773/401=-1135/401

    RonL
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Aug 2006
    Posts
    38
    Quote Originally Posted by CaptainBlack
    That is correct, now subtract the second equation fron this to get:

    420p - 19p + 14Q - 14Q = 770 + 3,

    simplifying:

    401p = 773,

    p = 773/401.

    Now going back to your first equation:

    30p + Q = 55,

    so:

    Q=55-30p=55-30.773/401=-1135/401

    RonL
    Hi can i know how you get
    =-1135/401
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by danielwu
    Quote Originally Posted by CaptainBlack
    ...

    Q=55-30p=55-30.773/401=-1135/401

    RonL
    Hi can i know how you get
    =-1135/401
    We have found that p=773/401.

    Q=55-30p is just a rearrangement of one of the equations that you were
    given.

    So:

    Q=55-30\times \frac{773}{401}.

    Now 55=\frac{55 \times 401}{401}=\frac{22055}{401}, so:

    Q=55-30\times \frac{773}{401}=\frac{22055}{401}-\frac{23190}{401}=-1135/401.

    RonL
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member
    Joined
    Aug 2006
    Posts
    38
    Hi was hoping someone could help me with this

    show that the given systems of equations have either an unlimited number of solutions or no solution. if there is an unlimited number of solutions, find one of them.

    <br />
x - 2y - 3z = 2<br />
    <br />
x - 4y -13z = 14<br />
    <br />
-3x + 5y +4z = 0<br />
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by danielwu
    Hi was hoping someone could help me with this

    show that the given systems of equations have either an unlimited number of solutions or no solution. if there is an unlimited number of solutions, find one of them.

    <br />
x - 2y - 3z = 2<br />
    <br />
x - 4y -13z = 14<br />
    <br />
-3x + 5y +4z = 0<br />
    You look at its determinant which is zero.
    Then you use Cramer's rule and see if each D_x,D_y,D_z is also zero or not.
    Since each is zero there system has infinitely many solutions.
    Attached Thumbnails Attached Thumbnails system of linear equations-picture3.gif  
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Junior Member
    Joined
    Aug 2006
    Posts
    38
    Quote Originally Posted by ThePerfectHacker
    You look at its determinant which is zero.
    Then you use Cramer's rule and see if each D_x,D_y,D_z is also zero or not.
    Since each is zero there system has infinitely many solutions.
    i have understand the cramer's rule.
    but this equation i have to solve it algebraically.
    thus i am really worst at algebra.

    can help me futher?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 3 123 LastLast

Similar Math Help Forum Discussions

  1. System of linear equations.
    Posted in the Algebra Forum
    Replies: 6
    Last Post: October 29th 2010, 04:48 PM
  2. Non-linear system of two equations
    Posted in the Algebra Forum
    Replies: 5
    Last Post: October 3rd 2010, 02:05 PM
  3. System of linear equations.
    Posted in the Algebra Forum
    Replies: 5
    Last Post: August 10th 2010, 12:03 AM
  4. Replies: 2
    Last Post: April 20th 2010, 04:26 PM
  5. System of non-linear equations
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 4th 2009, 09:18 PM

Search Tags


/mathhelpforum @mathhelpforum