solve the follwing equation
a/a^2-25+2/a+5=2/a-5
show all your work
To solve the equation a/a^2 - 25 + 2/a + 5 = 2/a - 5, we need to first find a common denominator for all the fractions on both sides of the equation.
The common denominator for the fractions a/a^2, 2/a, 2/a, and -5 is (a)(a + 5)(a - 5).
Therefore, the equation becomes:
a(a+5)(a-5)/(a)(a^2-25) + 2(a)(a-5)/(a)(a+5)(a-5) = 2(a)(a+5)/(a)(a-5),
Now we can simplify the equation further.
[(a^2 - 25)(a) + 2(a)(a-5)] / [(a)(a^2 - 25)] = 2(a)(a+5)/(a)(a-5)
(a^3 - 25a + 2a^2 - 10a) / (a)(a^2 - 25) = 2(a)(a+5)/(a)(a-5)
(a^3 + 2a^2 - 35a) / (a)(a^2 - 25) = 2(a)(a+5)/(a)(a-5)
Factoring the numerator of the left side:
a(a^2 + 2a - 35) / (a)(a+5)(a-5) = 2(a)(a+5)/(a)(a-5)
Next, we cancel out the common factors on both sides:
a/(a-5) = 2/(a+5)
a(a+5) = 2(a-5)
a^2 + 5a = 2a - 10
a^2 + 3a + 10 = 0
Unfortunately, this quadratic equation does not have real number solutions.