# Elimination method with 3 equations

• Apr 7th 2012, 11:18 PM
batosaims
Elimination method with 3 equations
http://i1209.photobucket.com/albums/...g?t=1333868898

I've tried this problem about 6 times and can't get the right answer in the book, can someone please show me how to do this...
• Apr 8th 2012, 01:03 AM
princeps
Re: Elimination method with 3 equations
Quote:

Originally Posted by batosaims
http://i1209.photobucket.com/albums/...g?t=1333868898

I've tried this problem about 6 times and can't get the right answer in the book, can someone please show me how to do this...

$\displaystyle \begin{cases}3a+5b-3c=-4 \\a-3b+c=6 \\-4a+6b+2c=-6\end{cases}$

Multiply 2nd. by -3 and add to the 1st. ,also..multiply 2nd. by 4 and add to the 3rd. :

$\displaystyle \begin{cases}a-3b+c=6 \\14b-6c=-22 \\-6b+6c=18\end{cases}$

Add 2nd to the 3rd. :

$\displaystyle \begin{cases}a-3b+c=6 \\14b-6c=-22\\8b=-4\end{cases}$

So , solution of the system is :

$\displaystyle (a,b,c)=\left(2,\frac{-1}{2},\frac{5}{2}\right)$
• Apr 8th 2012, 04:45 PM
Wilmer
Re: Elimination method with 3 equations
Quote:

Originally Posted by princeps
$\displaystyle \begin{cases}3a+5b-3c=-4 \\a-3b+c=6 \\-4a+6b+2c=-6\end{cases}$

Short cut; add 1st 2 equations to get:
4a + 2b - 2c = 2 ;add to 3rd equation:
-4a+ 6b +2c = -6
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8b = -4
b = -1/2 (as per Princeps)