1. 3/(x+2)+4/(x-5)=2/3
2. 5/(3x+8)=4/(3x-8)
I assume that by posting these equations, you are looking for solutions?
1. Multiply so everything is over the common denominator 3(x+2)(x-5)
3(x-5)3 + 4(x+2)3 = 2(x+2)(x-5)
all over the denominator. Now, the denominator doesn't really interest us, since it only makes sure that -2 or 5 cannot be one of the solutions. When the numerator is zero, then the fraction is zero.
9x-45 + 12x + 24 - 2x^2 + 6x +20 = 0
2x^2 + 27x -1 = 0
2. Place over a common denominator of (3x-8)(3x+8), so 8/3 and -8/3 cannot be solutions
5(3x-8) = 4(3x+8)
15x - 40 = 12x + 32
3x = 72
x=24