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-   -   Dependent Substitution and Elimination Problem! (https://www.askmehelpdesk.com/showthread.php?t=438371)

  • Jan 23, 2010, 07:56 PM
    luvUnionJack20
    Dependent Substitution and Elimination Problem!
    Hello!

    I was wondering if you'd be able to help me solve the following system of equations by elimination and substitution. I believe I know how to go about doing it, I'm just not sure what my answer would be...

    2x + 3y = 4
    -4x - 6y = -8

    Warning: When I finished working out the problem my two answers were 0 = 0 and
    -8 = -8.

    Thanks again!!
  • Jan 23, 2010, 08:42 PM
    Unknown008

    Ok, you must see something here.

    2x + 3y = 4
    3y = -2x + 4
    ----
    -4x - 6y = -8
    -6y = 4x-8
    3y = -2x+4

    The two equations are the same line! They were only in different 'forms'. Since solving two equations simultaneously means finding where they meet, well, there is only one answer, they meet all along the line 3y=-2x+4.
  • Jan 23, 2010, 08:46 PM
    luvUnionJack20
    Quote:

    Originally Posted by Unknown008 View Post
    Ok, you must see something here.

    2x + 3y = 4
    3y = -2x + 4
    ----
    -4x - 6y = -8
    -6y = 4x-8
    3y = -2x+4

    The two equations are the same exact line! They were only in different 'forms'. Since solving two equations simultaneously means finding where they meet, well, there is only one answer, they meet all along the line 3y=-2x+4.

    ah! So the answer for both elimination and substitution is 3y = -2x + 4? Would you need to completely isolate the y or would you leave it like that?
  • Jan 23, 2010, 08:48 PM
    Unknown008

    It does not matter. It's the same thing, except if we ask you to do so in the question.

    I hope it helped! :)
  • Jan 23, 2010, 09:02 PM
    luvUnionJack20
    Quote:

    Originally Posted by Unknown008 View Post
    It does not matter. It's the same thing, except if we ask you to do so in the question.

    I hope it helped! :)

    Mhmm! You really did! Thank you so much! :o)
  • Jan 23, 2010, 09:10 PM
    Unknown008

    You're welcome :)
  • Jan 23, 2010, 09:18 PM
    luvUnionJack20

    Uh... sorry to bother you again... just one quick question! Would you need to solve for the x? Sorry!
  • Jan 23, 2010, 09:27 PM
    Unknown008

    There is no need, since the line has different values of x.

    If there was one or more finite specific points, then you'll have to solve for x.

    If there was for example a domain (a specific region where the graph lies) then the point of solving for x arises. A most simple one is below;

    2x + 3y = 4
    -4x - 6y = -8 (for x>0)

    Then, your answer will be the line in question. However, the lines do not meet all the way along the line with equation 3y = -2x + 4, but only for values of x greater than 0. Your answer, then would be:

    3y = -2x + 4, for x > 0
  • Jan 23, 2010, 09:29 PM
    luvUnionJack20

    AH! That makes sense! Thanks again!
  • Jan 23, 2010, 11:50 PM
    morgaine300

    Just as an added note, if you simply solve it through like you already did and you get something where a = a and it's a true statement (0=0 and -8=-8 aren't solved but are true statements), it means the two lines are equal and on top of each other. You don't have to actually work out that they are equal.

    Also, if you come down to a false statement, then there's no solution and they are parallel lines. For instance:
    -4x + 2y = 8
    2x - y = 0
    using elimination solves down to 0=8, which is a false statement.

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