Given the equations: $a + b + c = 11$ $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c} = \frac{13}{17}$

We need to find: $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}$

Let us use the substitution $a = x, b = y, c = z$ for simplicity. Thus, we have: $x + y + z = 11$ $\frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} = \frac{13}{17}$

Next, consider the identity: $\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y}$

First, note that: $\frac{x}{y+z} = \frac{x}{11-x}$ $\frac{y}{z+x} = \frac{y}{11-y}$ $\frac{z}{x+y} = \frac{z}{11-z}$

Thus, we need to evaluate: $\frac{x}{11-x} + \frac{y}{11-y} + \frac{z}{11-z}$

To find this sum, we use a known result in algebra that for three variables $x, y, z$ such that $x + y + z = k$ and $\frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} = m$, the value of $\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y}$ can be found using:

$\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} = \frac{3}{2} \left( 1 + \frac{m}{k} \right)$

Substituting the given values $k = 11$ and $m = \frac{13}{17}$: $\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} = \frac{3}{2} \left( 1 + \frac{\frac{13}{17}}{11} \right)$ $= \frac{3}{2} \left( 1 + \frac{13}{187} \right)$ $= \frac{3}{2} \left( 1 + \frac{13}{187} \right)$ $= \frac{3}{2} \left( \frac{187 + 13}{187} \right)$ $= \frac{3}{2} \left( \frac{200}{187} \right)$ $= \frac{3}{2} \cdot \frac{200}{187}$ $= \frac{600}{374}$ $= \frac{300}{187}$

So, the value is: $\boxed{\frac{300}{187}}$

Would you like any details or have any questions?

Here are 5 questions you might be interested in:

1. How can I solve systems of equations involving three variables?
2. What are some common algebraic identities useful in problem-solving?
3. How do I approach solving complex fractions in equations?
4. Can you explain more about the identity used in this problem?
5. What techniques can help simplify algebraic expressions?

Tip: When working with systems of equations, always check for symmetry or known algebraic identities that might simplify the problem.