if a+b+c=11, and (1/[a+b])+(1/[b+c])+(1/[c+a])=(13/17), what is (a/[b+c])+(b/[c+a])+(c/[a+b]) equal to?

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

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

First, recall a useful identity in algebra which helps relate sums of fractions:

For any three numbers \(a, b, c\),
\[ \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = \frac{3}{2} \left( 1 + \frac{\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}}{a + b + c} \right) \]

Given:
\[ a + b + c = 11 \]
\[ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} = \frac{13}{17} \]

We substitute these values into the identity:

\[ \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = \frac{3}{2} \left( 1 + \frac{\frac{13}{17}}{11} \right) \]

First, simplify inside the parentheses:

\[ 1 + \frac{\frac{13}{17}}{11} = 1 + \frac{13}{17 \times 11} = 1 + \frac{13}{187} \]
\[ 1 + \frac{13}{187} = \frac{187}{187} + \frac{13}{187} = \frac{200}{187} \]

Now, substitute back into the formula:

\[ \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = \frac{3}{2} \left( \frac{200}{187} \right) \]
\[ = \frac{3 \times 200}{2 \times 187} = \frac{600}{374} = \frac{300}{187} \]

So, the final answer 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.

Learn how to solve an algebraic problem involving equations and fractions. Find (a/(b+c)) + (b/(c+a)) + (c/(a+b)) when given a + b + c = 11 and (1/(a+b)) + (1/(b+c)) + (1/(c+a)) = 13/17. This problem utilizes the sum of fractions identity and is suitable for Grades 10-12 students.

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