Question 117220
{{{((t-5)/(t-3))=((t-3)/(t+3))+(1/(t-2)))}}}


I'm just guessing, because you didn't specify, that you want to solve this equation for t.


First, put everything on the left:


{{{((t-5)/(t-3))-((t-3)/(t+3))-(1/(t-2)))=0}}}


Now you need to add the fractions.  All three have different binomial denominators, so you will need to multiply the three denominators together to get a common denominator:


{{{((t-5)(t+3)(t-2)/(t-3)(t+3)(t-2))-((t-3)(t-3)(t-2)/(t+3)(t-3)(t-2))-((t-3)(t+3)/(t-2)(t-3)(t+3))=0}}}


{{{(((t-5)(t+3)(t-2))-((t-3)(t-3)(t-2))-((t-3)(t+3)))/(t-2)(t-3)(t+3)=0}}}


Now you need to multiply all of the numerator factors together, and then combine like terms.  Leave the denominator alone because it won't help finding the function zeros, and we will be readily able to see the values excluded from the domain, namely 3, -3, and 2.


{{{((t^3-4t^2-11t+30)-(t^3-8t^2+21t-18)-(t^2-9))/((x-3)(x+3)(x-2))=0}}}


{{{(3t^2-32t+57)/((x-3)(x+3)(x-2))=0}}}


A rational function has zeros wherever the numerator has zeros, provided those values are in the domain of the overall rational function.  So use the quadratic formula to find the zeros of the numerator:


{{{x = (-(-32) +- sqrt( (-32)^2-4*3*57 ))/(2*3) =(32+-sqrt(340))/6=(16+-sqrt(85))/3}}}, roughly 8.4 and 2.3.  Neither root of the numerator is a root of the denominator, so both are valid solutions.


Hope that helps.
John