Simultaneous equations with three parts

Question

\begin{align*} 6a +24b +18c &= 168\\ 8a +28b +22c &= 208\\ 4a + 20b +20c &= 140 \end{align*} I've tried doing this multiplying so they cancelled out but I've always gotten decimal point answers. I know the answer but don't know how to get there $a= 10$, $b=3$, $c=2$.

Answer 1

The given equations are equivalent to: $$ \left\{\begin{array}{rcl}6a+24b+18c &=& 168\\ 8a+28b+22c&=&208\\ 12b+18c&=&72\end{array}\right.$$ or to:

$$ \left\{\begin{array}{rcl}6a+12b &=& 96\\ 2a+4b+4c&=&40\\ 12b+18c&=&72\end{array}\right.$$ that reduces to: $$ \left\{\begin{array}{rcl}a+2b &=& 16\\ a+2b+2c&=&20\\ 2b+3c&=&12\end{array}\right.$$ and subtracting the first equation from the second we get $c=2$, so $b=3$ and $a=10$ as wanted.

Answer 2

From the comment of KittyL your system become:

$$ \begin{cases} a+4b+3c=28\\ 4a+14b+11c=104\\ a+5b+5c=35 \end{cases} $$

Now: multiply the first equation by $4$ and subtract the second, then subtract the first equation from the third to obtain

$$ \begin{cases} 2b+c=8\\ b+2c=7\\ a+4b+3c=28 \end{cases} $$

Multiplying the second by $2$ and subtracting the first you have: $$ 3c=6 \Rightarrow c=2 \Rightarrow b= 7-4=3 \Rightarrow a=.... $$