My Math Forum Linear Systems of Equations: Substition Method

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 February 4th, 2010, 04:10 PM #1 Newbie   Joined: Feb 2010 Posts: 3 Thanks: 0 Linear Systems of Equations: Substition Method Why is that everytime I solve x in terms of y and substitute, I get a different answer (wrong)? I am trying to understand the intuition behind it.... ex: y=15-5x y=25-5x parallel lines means no unique solution substitute y in terms of x SUBSTITUTION 1: y=15-5x (15-5x)=25-5x 0=0 substitute x in terms of y SUBSTITUTION 2: y=15-5x therefore,5x=15-y y=25-(15-y) 2y=10 y=5 Makes no sense???
 February 4th, 2010, 06:42 PM #2 Newbie   Joined: Feb 2010 Posts: 3 Thanks: 0 Re: Linear Systems of Equations: Substition Method lol anyone???
 February 5th, 2010, 01:39 PM #3 Global Moderator   Joined: May 2007 Posts: 6,807 Thanks: 716 Re: Linear Systems of Equations: Substition Method The particular pair of equations you chose has no solution. You can easily see that by subtracting one from the other and get as an equation 15=25.
 February 7th, 2010, 04:57 AM #4 Newbie   Joined: Feb 2010 Posts: 3 Thanks: 0 Re: Linear Systems of Equations: Substition Method yes i am aware of this.... and you can prove it using substition... but my question is why is it that you can only solve systems of linear equations by substituing y in terms of x, and not x in terms of y? substitute y in terms of x SUBSTITUTION 1: y=15-5x (15-5x)=25-5x 0=0 substitute x in terms of y SUBSTITUTION 2: y=15-5x therefore,5x=15-y y=25-(15-y) 2y=10 y=5
 February 7th, 2010, 05:57 AM #5 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Linear Systems of Equations: Substition Method As long as there is a solution, you can always solve for either variable first. You have made an error in your second substitution: $y=25-(15-y)\quad\Rightarrow\quad y=25-15+y\quad\Rightarrow\quad 0=10,$ which is clearly false

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