Solve: \( \frac{3}{b-7}+\frac{4}{b+7}=\frac{42}{b^{2}-49} \) \( b=\square \) (Enter your answers, separated by commas, or type DNE if no solution exists)

To solve the equation \( \frac{3}{b-7}+\frac{4}{b+7}=\frac{42}{b^{2}-49} \), we first notice that \( b^2 - 49 \) can be factored as \( (b-7)(b+7) \). Therefore, we can rewrite the right-hand side of the equation: \[ \frac{3}{b-7}+\frac{4}{b+7}=\frac{42}{(b-7)(b+7)} \] Next, we can combine the fractions on the left side: The common denominator for the fractions on the left side is \( (b-7)(b+7) \). Thus, we rewrite it: \[ \frac{3(b+7) + 4(b-7)}{(b-7)(b+7)} = \frac{42}{(b-7)(b+7)} \] Now, we simplify the numerator: \[ 3(b+7) + 4(b-7) = 3b + 21 + 4b - 28 = 7b - 7 \] This allows us to rewrite the equation as: \[ \frac{7b - 7}{(b-7)(b+7)} = \frac{42}{(b-7)(b+7)} \] Since the denominators are the same (as long as \( b \neq 7 \) and \( b \neq -7 \)), we can equate the numerators: \[ 7b - 7 = 42 \] Now, we solve for \( b \): \[ 7b = 42 + 7 \] \[ 7b = 49 \] \[ b = 7 \] However, we need to remember that \( b \neq 7 \) or \( b \neq -7 \) because of the restrictions from the original equation. So \( b = 7 \) does not satisfy the equation. Now let’s check \( b = -7 \): Substituting \( b = -7 \): \[ \frac{3}{-7-7}+\frac{4}{-7+7} \quad \text{(undefined)} \] Again, \( b = -7 \) is also not a solution. Since both potential solutions are not valid due to the restrictions we found, we conclude: There is no solution. Final answer: DNE