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...
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 :
$\displaystyle (a,b,c)=\left(2,\frac{-1}{2},\frac{5}{2}\right)$