\[ \frac{16-u^{2}}{4 u^{2}-16 u}=-\frac{4+u}{4 u} \] Part: \( 1 / 2 \) Part 2 of 2 If there is more than one restriction, use the "and" button. There are restrictions on the variable \( u \). There are no restrictions on the variable \( u \).

To solve the given equation and identify restrictions on the variable \( u \), let’s first analyze each side of the equation. Rewriting the original equation: \[ \frac{16 - u^2}{4u^2 - 16u} = -\frac{4 + u}{4u} \] **First additional information:** One of the key restrictions comes from the denominators of the fractions. The expression \( 4u^2 - 16u \) can be factored as \( 4u(u - 4) \), which means \( u \) cannot be 0 or 4 because you can’t divide by zero! Additionally, in the negative fraction, the term \( 4u \) also means \( u \) cannot be 0. Thus, the restrictions on \( u \) are \( u \neq 0 \) and \( u \neq 4 \). **Second additional information:** Sometimes, complex equations like this can seem daunting, but always start simplifying where you can! After determining the restrictions, it’s wise to cross-multiply to eliminate the fractions (just be cautious about your restrictions). Remember to check your final answers against your restrictions to ensure you didn’t accidentally reintroduce division by zero. This tactic makes problem-solving straight-forward and helps avoid common mistakes!