Results 1 to 5 of 5

Math Help - linear equations

  1. #1
    Junior Member
    Joined
    Sep 2006
    Posts
    38

    linear equations

    can anyone explain how to slove this? there's 3 equations and 3 variables, but only 2 variables in each equation
    2a-5c=-18
    7b-8c=-65
    3a-2b=2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by imthatgirl View Post
    can anyone explain how to slove this? there's 3 equations and 3 variables, but only 2 variables in each equation
    2a-5c=-18
    7b-8c=-65
    3a-2b=2
    you have: 2a-5c=-18

    then: 2a=5c-18

    thus: a=(5/2)c-9

    You also have equation: 3a-2b=2

    substitute: 3((5/2)c-9)-2b=2

    then: (15/2)c-27-2b=2

    then: -2b=2+27-(15/2)c

    then: b=(15/4)c-(27/2)-1

    You also have equation: 7b-8c=-65

    substitute: 7((15/4)c-(27/2)-1)-8c=-65

    then: (105/4)c-(189/2)-8c=-65

    now you have an equation with one variable...
    make sure to check my arithmetic
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member classicstrings's Avatar
    Joined
    Mar 2006
    Posts
    175
    Awards
    1
    Quote Originally Posted by imthatgirl View Post
    can anyone explain how to slove this? there's 3 equations and 3 variables, but only 2 variables in each equation
    2a-5c=-18
    7b-8c=-65
    3a-2b=2
    3 equations, 3 unknowns = solvable.
    3 equations, 4 unknowns = a system of equations* (check) i.e. many solutions

    4 equations, 4 unknowns = solvable
    ... and so on
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,710
    Thanks
    630
    Hello, imthatgirl!

    Can anyone explain how to solve this?
    There's 3 equations and 3 variables, but only 2 variables in each equation.

    . . 2a - 5c .= .-18
    . . 7b - 8c .= .-65
    . . 3a - 2b .= . 2

    Missing variables should not be a problem.
    If you're using Elimination, they actually did us a favor!
    Code:
          (1)   2a      - 5c  =  -18
          (2)        7b - 8c  =  -65
          (3)   3a - 2b       =    2


    Multiply (1) by -3, multiply (3) by 2, and add:
    Code:
    -3(1)      -6a      + 15c  =  54
     2(3)       6a - 4b        =   4
                - - - - - - - - - - -
          (4)      - 4b + 15c  =  58

    Multiply (4) by 7, multiply (2) by 4, and add:
    Code:
    7(4)     -28b + 105c  =  406
    4(2)      28b -  32c  = -260
             - - - - - - - - - -
                     73c  =  146

    And we have: .c = 2

    Substitute into (1): . 2a - 52 .= .-18 . . a = -4

    Substitute into (2): . 7b - 82 .= .-65 . . b = -7


    Therefore: .(a,b,c) .= .(-4,-7,2)

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by classicstrings View Post
    3 equations, 3 unknowns = solvable.
    3 equations, 4 unknowns = a system of equations* (check) i.e. many solutions

    4 equations, 4 unknowns = solvable
    ... and so on
    Not always.
    ---
    Firstly, we for simplicity sake are speaking about linear equations. [Otherwise, the mathematics that studies ploynomial systems is called algebraic geometry, we are speaking about linear algebra-meaning linear systems].

    The necessary and sufficient condition for a n variable n equations system have a unique solution is that the determinant is non-zero.

    For example,
    x+y=1
    x+y=2

    Has the solution set as an empty set.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: November 30th 2011, 01:41 AM
  2. Linear Alegebra linear equations proofs
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: January 13th 2010, 09:47 AM
  3. Replies: 7
    Last Post: August 30th 2009, 10:03 AM
  4. Replies: 3
    Last Post: February 27th 2009, 07:05 PM
  5. Replies: 1
    Last Post: July 29th 2007, 02:37 PM

Search Tags


/mathhelpforum @mathhelpforum