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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

## #4 2005-07-26 20:47:52

schwebba
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### Re: Substitution Method (where two lines cross)

wow thanks i think i get it now. you sure cleared things up a bit

gonna practise some more on this..

thanks a lot for that nice solution

## #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