Help solve the algebra equation below ?

.....1.............2................2

--------- +  ---------- = -------------------]

.n - 4...........n - 2.........n^2 - 6n + 8

....1.............2................2 

[-------- + ----------- = -------------------- ](n - 4)(n - 2)

..n - 4.......n - 2...........(n - 4)(n - 2)

 n - 2 + n - 4 = 2

2n - 6 = 2

2n = 2 + 6

2n = 8

 n = 4 Answer//

n^2 - 6n + 8 = (n-4)(n-2)

so IF n not = 4 and if n not = 2

then n-2 + 2(n-4) = 2

==> 3n -2 -8 = 2

==> 3n = 12

==> n = 4

so, no solution, since n can't be 4

= n² - 6n + 8

= n² - (4n + 2n) + 8

= n² - 4n - 2n + 8

= (n² - 4n) - (2n - 8)

= n.(n - 4) - 2.(n - 4)

= (n - 4).(n - 2)

[1/(n - 4)] + (2/(n - 2)] = 2/(n² - 6n + 8)

[1/(n - 4)] + (2/(n - 2)] = 2/[(n - 4).(n - 2)] → where: n ≠ 4 and where: n ≠ 2

[(n - 2) + 2.(n - 4)]/[(n - 4).(n - 2)] = 2/[(n - 4).(n - 2)] → you can simplify

(n - 2) + 2.(n - 4) = 2

n - 2 + 2n - 8 = 2

3n = 12

n = 4 ← no possible because the condition

No solution

(1/(n - 4)) + (2/(n - 2)) = 2/(n^2 - 6n + 8)

All rational expressions need a common denominator; since (n - 4)(n - 2) = n^2 - 6n + 8, the right side can stay as is.  Multiply the first expression on the left by (n - 2)/(n - 2) and the second by (n - 4)/(n - 4).

(1/(n - 4))((n - 2)/(n - 2)) + (2/(n - 2))((n - 4)/(n - 4)) = 2/(n^2 - 6n + 8)

((n - 2) / (n^2 - 6n + 8)) + ((2(n - 4)) / (n^2 - 6n + 8)) = 2/(n^2 - 6n + 8)

You now have like denominators, so dispose of them and deal with the numerators.

(n - 2) + 2(n - 4) = 2

(n - 2) + (2n - 8) = 2

3n - 10 = 2

3n = 12

n = 4.  Except...

...this solution results in division by 0 on the right and in 1/(n - 4) on the left.  This makes n = 4 extraneous and thus you have no solutions possible.

   1          2              2

------ + ------- = --------------

n - 4     n - 2     n² - 6n + 8

(n-2) + 2(n-4)           2

------------------ = ---------------   ⇒

 n² - 6n + 8       n² - 6n + 8

(n-2) + 2(n-4) = 2

n - 2 + 2n - 8 = 2

3n - 10 = 2

3n = 12

n = 4 ....................ANS

 noticed (n-4)(n-2) = n² -6n +8

And, to add fractions you must have a common denominator

Multiply by missing factor:

1/(n-4) • (n-2)/(n-2) = (n-2)/(n² -6n +8)

2/(n-2) • (n-4)/(n-4) = 2(n-4)/(n² -6n +8)

Since all the terms have (n² -6n +8) in denominator, it can be elimated leaving:

(n-2) + 2(n-4) = 1

Gather all like terms:

3n -10 = 1

3n = 11

n = 11/3

1 / (n - 4) + 2 / (n - 2) = 2 / (n² - 6n + 8)

The first thing that I'll do is factor the right side's denominator so we can determine what the LCD is for this equation:

1 / (n - 4) + 2 / (n - 2) = 2 / [(n - 4)(x - 2)]

The LCD here is (n - 4)(n - 2) which means that n cannot be equal to 2 or 4, otherwise we will be dividing by zero.

Now we can multiply both sides by this LCD to get rid of the fractions:

(n - 2) + 2(n - 4) = 2

Expand the left side:

n - 2 + 2n - 8 = 2

3n - 10 = 2

Now we can finish solving for n:

3n = 12

n = 4

But since 4 is one of the values we said upfront that n cannot be, the final answer is that there is no real solution to this equation.