=(x-5)/(x-5)+ 10/(x-5)
= 1+ 10/(x-5)
Similarly: (x-6)/(x+6) =1-12/(x+6)
(x+4)/(x-4)= 1+8/(x-4)
(x-7)/(x+7)= 1- 14/(x+7)
So the original equation is:
[ 1+ 10/(x-5)]-[ 1- 12/(x+6)]
10/(x-5)+ 12/(x+6)= 8/(x-4)+ 14/(x+7)
5/(x-5)+6/(x+6)=4/(x-4)+7/(x+7)
Then separate the two sides of the equation.
[5(x+6)+6(x-5)]/(x-5)(x+6)=[4(x+7)+(7(x-4)]/(x-4)(x+7)
1 1x/(x^2+x-30)= 1 1x/(x^2+3x-28)
1 1x(x^2+3x-28)= 1 1x(x^2+x-30)
1 1x(x^2+3x-28)- 1 1x(x^2+x-30)=0
1 1x(x^2+3x-28-x^2-x+30)=0
1 1x(2x+2)=0
x(x+ 1)=0
So x 1=0, x2=- 1.
It is verified that X 1 = 0 and X2 =- 1 are the solutions of the original fractional equation.
Do you understand?