I'm having trouble with +/- when solving for y

lostgirl

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Jan 27, 2008
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I have an example from the book, but I don't quite understand it.
4x-6y=12 I need to get it into y=Mx+b form so that I can graph it, but I get lost between steps 1 and 2. I know I should re-arrange the equation, but why do I switch the +/- signs?

These are the steps as they appear in the book:
4x-6y=12
step 1) 4x=6y+12 ( the sign changes from + to - )
step 2) 6y=4x - 12
step 3) 6y=4x-12
step 4) y=(4/6)x-12/6
solution: y=(2/3)x-2
could you explain the steps to me and why I do what I do in each step. ie:switching signs
 

o_O

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Oct 20, 2007
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Step 1: 4x = 6y + 12

What happens from Step 1 to Step 2 is that you subtracted 12 from both sides to isolate y by itself:

4x - 12 = 6y + 12 - 12
4x - 12 = 6y + 0
6y = 4x - 12 (a - b = c is the same thing as saying c = a - b if that is what's confusing you)
 

stapel

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Feb 4, 2004
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lostgirl said:
I get lost between steps 1 and 2. I know I should re-arrange the equation, but why do I switch the +/- signs?
Many lessons on solving linear equations show the computations (and the reasoning behind them) in more detail than it appears is exhibited in your text. Try studying a few of the online offerings for further information and worked examples. :idea:

Have fun! :D

Eliz.
 

lostgirl

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Jan 27, 2008
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Thanks sooooo much!
I believe I've got it now. :D
 
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