can you help me solve this math equation?

Question

You're a student in an Algebra class; a friend emails you asking for help on a problem. Explain the process you take to solve the problem:

x-2 7 3
--------- = ------ + -------
x^2-25 x+5 x - 5

your friend is the type of student that needs to kno why things are being done. give specific reasons for each step

Answer 1

So a/(x+5) +b/(x-5)=x-273/(x^2-25)
or a(x-5)+b(x+5)=x-273
so a+b=1 and 5b-5a=-273
Solving the first of these equations, a=1-b: substitute this into the second equation. 5b-5(1-b)=-273
or 10b-5=-273
or 10b=-268
so b=-26.8 and a = 27.8

Answer 2

Your graphics didn't seem to come through very clear. I'm assuming your equation is

[(x-2)/(x^2-25)] = [7/(x+5)] + [3/(x-5)].

If this is the case, then you have the difference of two squares on the left side so we can get common denominators on the right by multiplying each term by the opposite denominator. Thus,

[(x-2)/(x^2-25)] = [7(x-5)/(x+5)(x-5)] + [3(x+5)/(x-5)(x+5)].

Note, that we haven't changed the value of the terms because (x-5)/(x-5) and (x+5)/(x+5) both equal one. Simplifying the right side gives us

[(x-2)/(x^2-25)] = [(7x-35)/(x^2-25)] + [(3x+15)/(x^2-25)]
[(x-2)/(x^2-25)] = [(10x-20)/(x^2-25)]

Multiplying both sides by x^2-25 to clear the fractions leaves us with

x-2=10x-20 Add 20 to and subtract x from both sides.
18=9x Divide both sides by 9.
2=x.

Answer 3

Next time use / when you want to divide, not --------

Partial fractions
(x-2) / (x^2 - 25) = (x-2) / (x+5)(x-5)
= A/(x+5) + B/(x-5)
= [ A*(x-5) + B*(x+5)] / (x-5)(x+5)
Numerator = (A+B)x + 5*(B-A) = x-2
So A+B = 1
B-A = -2/5

B = 3/10
A = 7/10