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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...

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- Apr 7th 2012, 11:18 PMbatosaimsElimination method with 3 equations
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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 AMprincepsRe: Elimination method with 3 equations
$\displaystyle \begin{cases}3a+5b-3c=-4 \\a-3b+c=6 \\-4a+6b+2c=-6\end{cases}$

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 PMWilmerRe: Elimination method with 3 equations