<s>[INST] Ratio and proportion problem 
Question:

If $$(a+b):(b+c):(c+a)=6:7:8$$ and $a+b+c=14$, then find the value of $c$.


My solution:


*

*$$\frac{(a+b)(c+a)}{(b+c)}=\frac{(6)(8)}{7}$$$$\Rightarrow \frac{ac + a^2 + bc + ba}{b+c} = \frac{48}{7}$$$$\Rightarrow \frac{a(b+c)+a^2+bc}{b+c}=\frac{48}{7}$$$$\Rightarrow ????$$

*$$\Rightarrow a+b=6x \space\space \text{and} \space \space b+c=7x$$$$\Rightarrow b=6x-a\space\space\space\text{and}\space\space\space b=7x-c$$$$\Rightarrow \text{solving we get}\space\space x = c-a$$$$\Rightarrow ????$$

My query:
I am totally stuck on this problem. Please help. 
Thanks a lot!
 [/INST] Let $$x=b+c,$$ $$y=a+c,$$ $$z=a+b.$$ Then,  $x+y+z=2(a+b+c)=28$ and $z:x:y=6:7:8$. 
By substituting $z=6k$, $x=7k$, $y=8k$ into $x+y+z=28$, we have $21 k=28$, $k=\frac{4}{3}$, from where we get: $$z=6k=8,$$ $$x=7k=\frac{28}{3},$$ $$y=8k=\frac{32}{3}.$$
Finally, $$a=a+b+c-x=\frac{14}{3},$$ $$b=a+b+c-y=\frac{10}{3},$$ $$c=a+b+c-z=6.$$
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