Question 633083
{{{((n-3))/((6n-6))}}} = {{{1/9}}} - {{{((n-3))/((4n-4))}}}
We can do some factoring here
{{{((n-3))/(6(n-1))}}} = {{{1/9}}} - {{{((n-3))/(4(n-1))}}}
Right here we can answer the question about excluded values
You can see that n cannot = 1, as it would put zeros in the denominators
:
The common denominator would be 36(n-1), multiply equation by this
36(n-1)*{{{((n-3))/(6(n-1))}}} = 36(n-1)*{{{1/9}}} - 36(n-1)*{{{((n-3))/(4(n-1))}}}
Cancel the denominators and you have
6(n-3) = 4(n-1) - 9(n-3)
6n - 18 = 4n - 4 - 9n + 27
6n - 18 = -5n + 23
Combine n's on the left, numbers on the right
6n + 5n = 23 + 18
11n = 41
n = {{{41/11}}}
n ~ 3.73, kind of a nasty solution but it seems to check out. C