1/(x-5)-4/(x+2)=7/(x^2-3x-10)

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Solution for 1/(x-5)-4/(x+2)=7/(x^2-3x-10) equation: 


D( x )

x-5 = 0

x^2-(3*x)-10 = 0

x+2 = 0

x-5 = 0

x-5 = 0

x-5 = 0 // + 5

x = 5

x^2-(3*x)-10 = 0

x^2-(3*x)-10 = 0

x^2-3*x-10 = 0

x^2-3*x-10 = 0

DELTA = (-3)^2-(-10*1*4)

DELTA = 49

DELTA > 0

x = (49^(1/2)+3)/(1*2) or x = (3-49^(1/2))/(1*2)

x = 5 or x = -2

x+2 = 0

x+2 = 0

x+2 = 0 // - 2

x = -2

x in (-oo:-2) U (-2:5) U (5:+oo)

1/(x-5)-(4/(x+2)) = 7/(x^2-(3*x)-10) // - 7/(x^2-(3*x)-10)

1/(x-5)-(4/(x+2))-(7/(x^2-(3*x)-10)) = 0

1/(x-5)-4*(x+2)^-1-7*(x^2-3*x-10)^-1 = 0

1/(x-5)-4/(x+2)-7/(x^2-3*x-10) = 0

x^2-3*x-10 = 0

x^2-3*x-10 = 0

x^2-3*x-10 = 0

DELTA = (-3)^2-(-10*1*4)

DELTA = 49

DELTA > 0

x = (49^(1/2)+3)/(1*2) or x = (3-49^(1/2))/(1*2)

x = 5 or x = -2

(x+2)*(x-5) = 0

1/(x-5)-4/(x+2)-7/((x+2)*(x-5)) = 0

(1*(x+2))/((x-5)*(x+2))+(-4*(x-5))/((x-5)*(x+2))-7/((x-5)*(x+2)) = 0

1*(x+2)-4*(x-5)-7 = 0

22-3*x-7 = 0

15-3*x = 0

(15-3*x)/((x-5)*(x+2)) = 0

(15-3*x)/((x-5)*(x+2)) = 0 // * (x-5)*(x+2)

15-3*x = 0

15-3*x = 0 // - 15

-3*x = -15 // : -3

x = -15/(-3)

x = 5

x in { 5} 

x belongs to the empty set

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