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February 4th, 2010, 04:10 PM   #1
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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???
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February 4th, 2010, 06:42 PM   #2
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Re: Linear Systems of Equations: Substition Method

lol anyone???
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February 5th, 2010, 01:39 PM   #3
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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.
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February 7th, 2010, 04:57 AM   #4
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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
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February 7th, 2010, 05:57 AM   #5
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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:

which is clearly false
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