Question: Determine if the following linear equation has one solution, no solution, or if the solution is all real numbers. Express your solution as either a simplified fraction or decimal rounded to two places.
$-8 - 8(4n - 3) = -8(-n - 3) + 6$
1. **State the problem:** We need to determine if the linear equation $$-8 - 8(4n - 3) = -8(-n - 3) + 6$$ has one solution, no solution, or infinitely many solutions (all real numbers).
2. **Expand both sides:**
$$-8 - 8(4n - 3) = -8(-n - 3) + 6$$
Distribute the $-8$ on the left:
$$-8 - 8 \times 4n + 8 \times 3 = -8(-n - 3) + 6$$
$$-8 - 32n + 24 = -8(-n - 3) + 6$$
Simplify the left side:
$$(-8 + 24) - 32n = -8(-n - 3) + 6$$
$$16 - 32n = -8(-n - 3) + 6$$
3. **Expand the right side:**
$$-8(-n - 3) = -8 \times (-n) - 8 \times (-3) = 8n + 24$$
So the equation becomes:
$$16 - 32n = 8n + 24 + 6$$
$$16 - 32n = 8n + 30$$
4. **Collect like terms:**
Bring all terms involving $n$ to one side and constants to the other:
$$16 - 32n = 8n + 30$$
Add $32n$ to both sides:
$$16 = 8n + 30 + 32n$$
$$16 = 40n + 30$$
Subtract $30$ from both sides:
$$16 - 30 = 40n$$
$$-14 = 40n$$
5. **Solve for $n$:**
Divide both sides by $40$:
$$n = \frac{-14}{40}$$
Simplify the fraction by dividing numerator and denominator by $2$:
$$n = \frac{\cancel{-14}^{7}}{\cancel{40}^{20}} = -\frac{7}{20}$$
6. **Convert to decimal rounded to two places:**
$$n = -0.35$$
7. **Conclusion:** The equation has one solution, $$n = -\frac{7}{20}$$ or $$n = -0.35$$.
**Answer:** One Solution $n = -\frac{7}{20}$