Question

1. (\frac{3}{2n} + \frac{1}{2n^2 - 2n} = \frac{n - 4}{2n^2 - 2n})

Explanation:

Step1: Factor denominators

Notice that $2n^2-2n = 2n(n-1)$, so rewrite the equation:
$$\frac{3}{2n} + \frac{1}{2n(n-1)} = \frac{n-4}{2n(n-1)}$$

Step2: Eliminate denominators

Multiply all terms by $2n(n-1)$ (where $n
eq0,1$):
$$3(n-1) + 1 = n-4$$

Step3: Expand left-hand side

$$3n - 3 + 1 = n-4$$

Step4: Simplify left-hand side

$$3n - 2 = n-4$$

Step5: Isolate variable terms

Subtract $n$ from both sides:
$$2n - 2 = -4$$

Step6: Isolate constant terms

Add 2 to both sides:
$$2n = -2$$

Step7: Solve for n

Divide by 2:
$$n = \frac{-2}{2} = -1$$

Step8: Verify solution

Substitute $n=-1$ into original equation:
Left: $\frac{3}{2(-1)} + \frac{1}{2(-1)^2-2(-1)} = -\frac{3}{2} + \frac{1}{2+2} = -\frac{3}{2} + \frac{1}{4} = -\frac{5}{4}$
Right: $\frac{-1-4}{2(-1)^2-2(-1)} = \frac{-5}{2+2} = -\frac{5}{4}$
Both sides are equal, so solution is valid.

Answer:

$n=-1$