Solve: $$ \frac{n}{n-4}=-\frac{32}{7} $$ . Enter the exact value of $$ n $$ as a fraction if necessary.

Question

UpStudy AI · Accepted Answer

Multiply both sides of the equation by $$(n-4)$$ to eliminate the fraction:

$$
\frac{n}{n-4} \times (n-4) = -\frac{32}{7} \times (n-4)
$$

This simplifies to:

$$
n = -\frac{32}{7}(n-4)
$$

Distribute $$-\frac{32}{7}$$ on the right-hand side:

$$
n = -\frac{32}{7} \, n + \frac{128}{7}
$$

Add $$\frac{32}{7} \, n$$ to both sides to combine like terms:

$$
n + \frac{32}{7} \, n = \frac{128}{7}
$$

Rewrite the left-hand side with a common denominator:

$$
\frac{7n + 32n}{7} = \frac{128}{7}
$$

Simplify the numerator:

$$
\frac{39n}{7} = \frac{128}{7}
$$

Multiply both sides by 7 to eliminate the denominator:

$$
39n = 128
$$

Finally, solve for $$n$$ by dividing both sides by 39:

$$
n = \frac{128}{39}
$$
$$ n = \frac{128}{39} $$

Solve: \( \frac{n}{n-4}=-\frac{32}{7} \) . Enter the exact value of \( n \) as a fraction if necessary.