How do I solve this problem? 
Exercise: If $a+2b=125$ and $b+c=348$, find out $2a+7b+3c$. Here $a$, $b$, $c$ are natural numbers.

The answer is: $2a+7b+3c = 1294$
I tried but just can't figure out how to get to this answer. I have a lot of exercises similar to this one but don't know how to aproach them. Can anyone write the steps in order to get to the answer above? Also  it would be great if you can write in a general way so I can apply it to other exercises similar to this.
 A: Hint: $2a + 7b + 3c = 2(a+2b) + 3(b+c)$.
A: Approach that often works:
You want to have some expression for $2a+7b+3c$, you know $a+2b=125$ and $b+c=384$.
Note that $a$ only occurs in the first expression, so you have to subtract it $2$ times from your expression.
What remains is
$2a+7b+3c-2*(a+2b)=3b+3c$
Now for this exercise to be solvable you have to express $3b+3c$ with $b+c$, which is easy. You will end up with the form Cocopuffs gave you.
A: If you multiply the first equation of the following system by $2$ and the second equation by $3$ you get an equivalent system.
$$\left\{ 
\begin{array}{c}
a+2b=125 \\ 
b+c=348
\end{array}
\right. \overset{\times 2}{\underset{\times 3}{\Leftrightarrow }}\left\{ 
\begin{array}{c}
2a+4b=250 \\ 
3b+3c=1044
\end{array}
\right. $$
Now you can add the two equations ...
A: Hint $\quad\begin{eqnarray}\rm\ j\,(a\!+\!2b) + k\,(b\!+\!c) &=&\,\rm j\,a + (2\,j\!+\!k)\,b + k\, c \\ 
&=&\,\rm 2\,a\ \ \ +\ \ \ 7\ b\ \ +\ \ 3\,c\ \ \Rightarrow\ \  j,k = \ldots \end{eqnarray}$
