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#1 2005-07-26 10:29:53
- schwebba
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Substitution Method (where two lines cross)
I got some trouble with finding out how to calculate this..
Question: Where do these lines cross eachother x + 3y +3=0 and x - 3y +2 =0
First i tried to solve y and got this far
y = -1/3x - 1 = 1/3x+2/3 (divided all with 3)
but im not even sure if that is right. can anyone please explain how i can do this?
Last edited by schwebba (2005-07-26 10:32:03)
#2 2005-07-26 14:26:05
- ganesh
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Re: Substitution Method (where two lines cross)
Solving the two equations, you get
x = -5/2, y = -1/6
Hence, the two lines cross at
(-5/2, -1/6)
How the two equations are solved:-
First equation can be written as x+3y = -3
The second equation can be written as x - 3y = -2
Adding the two, we get 2x = -5, or x = -5/2
Substituting this value in equation (1) -5/2 + 3y = -3
3y = -3 + 5/2
3y = -1/2
y = -1/6
Character is who you are when no one is looking.
#3 2005-07-26 17:10:46
- MathsIsFun
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Re: Substitution Method (where two lines cross)
Neatly solved!
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#5 2005-07-28 19:09:04
- MathsIsFun
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Re: Substitution Method (where two lines cross)
And for all those avid graphers out there, I have a little tool to figure out lines from points here: http://www.mathsisfun.com/straight-line … ulate.html
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman