$$ \frac{16-u^{2}}{4 u^{2}-16 u}=-\frac{4+u}{4 u} $$ 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 $$ u $$.

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Solve the equation $$ \frac{16-u^{2}}{4 u^{2}-16 u}=-\frac{4+u}{4 u} $$.
Solve the equation by following steps:
- step0: Solve for $$u$$:
  $$\frac{16-u^{2}}{4u^{2}-16u}=-\frac{4+u}{4u}$$
- step1: Find the domain:
  $$\frac{16-u^{2}}{4u^{2}-16u}=-\frac{4+u}{4u},u \in \left(-\infty,0\right)\cup \left(0,4\right)\cup \left(4,+\infty\right)$$
- step2: Divide the terms:
  $$-\frac{4+u}{4u}=-\frac{4+u}{4u}$$
- step3: The statement is true:
  $$u \in \mathbb{R}$$
- step4: Check if the solution is in the defined range:
  $$u \in \mathbb{R},u \in \left(-\infty,0\right)\cup \left(0,4\right)\cup \left(4,+\infty\right)$$
- step5: Find the intersection:
  $$u \in \left(-\infty,0\right)\cup \left(0,4\right)\cup \left(4,+\infty\right)$$
The solution to the equation is $$ u \in (-\infty,0) \cup (0,4) \cup (4,+\infty) $$.
The solution is $$ u $$ is any real number except 0 and 4.

Part:

Part 2 of 2

If there is more than one restriction, use the “and” button.
There are restrictions on the variable .
There are no restrictions on the variable .

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Part: Part 2 of 2 If there is more than one restriction, use the “and” button. There are restrictions on the variable . There are no restrictions on the variable .