If \( \frac{a}{b+c}=\frac{1}{3}, \frac{c}{a+b}=\frac{5}{7} \), Find the value of: \( \frac{b}{a+c} \)

Given the equations \[ \frac{a}{b+c}=\frac{1}{3} \] and \[ \frac{c}{a+b}=\frac{5}{7}, \] we can express \( a \) and \( c \) in terms of \( b \) and constants. From the first equation, we can rearrange it as follows: \[ 3a = b + c \quad \text{(1)} \] From the second equation, we have: \[ 7c = 5(a + b) \quad \text{(2)} \] Substituting \( b + c \) from equation (1) into equation (2): Using equation (1), we can express \( c \) as \( c = 3a - b \). Substitute this into equation (2): \[ 7(3a - b) = 5(a + b). \] Expanding both sides: \[ 21a - 7b = 5a + 5b. \] Rearranging terms gives us: \[ 21a - 5a = 7b + 5b, \] or \[ 16a = 12b. \] Thus, we can simplify this to: \[ \frac{a}{b} = \frac{12}{16} = \frac{3}{4}, \quad \text{so } a = \frac{3}{4}b. \] Now, substitute \( a \) back into equation (1): \[ 3\left(\frac{3}{4}b\right) = b + c. \] This simplifies to: \[ \frac{9}{4}b = b + c. \] Rearranging gives us: \[ c = \frac{9}{4}b - b = \frac{9}{4}b - \frac{4}{4}b = \frac{5}{4}b. \] Now we have: \[ a = \frac{3}{4}b, \quad c = \frac{5}{4}b. \] Next, let’s find \( \frac{b}{a+c} \): \[ a + c = \frac{3}{4}b + \frac{5}{4}b = \frac{8}{4}b = 2b. \] Thus, \[ \frac{b}{a+c} = \frac{b}{2b} = \frac{1}{2}. \] Hence, the value of \( \frac{b}{a+c} \) is \[ \boxed{\frac{1}{2}}. \]