Problem 9.4. For positive numbers $a, b, c$, it is known that

$$
\frac{a+b+c}{a+b-c}=7, \quad \frac{a+b+c}{a+c-b}=1.75
$$

What is the value of $\frac{a+b+c}{b+c-a}$?
Answer: 3.5.

Solution. Let $\frac{a+b+c}{b+c-a}=x$. Notice that

$$
\begin{gathered}
\frac{1}{7}+\frac{1}{1.75}+\frac{1}{x}=\frac{a+b-c}{a+b+c}+\frac{a+c-b}{a+b+c}+\frac{b+c-a}{a+b+c}=1 \\
\frac{1}{7}+\frac{4}{7}+\frac{1}{x}=1 \\
x=3.5
\end{gathered}
$$