(7/(x^2-x-12))+(4/(x+3))=(1/(x-4))

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Solution for (7/(x^2-x-12))+(4/(x+3))=(1/(x-4)) equation: 


D( x )

x^2-x-12 = 0

x-4 = 0

x+3 = 0

x^2-x-12 = 0

x^2-x-12 = 0

x^2-x-12 = 0

DELTA = (-1)^2-(-12*1*4)

DELTA = 49

DELTA > 0

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

x = 4 or x = -3

x-4 = 0

x-4 = 0

x-4 = 0 // + 4

x = 4

x+3 = 0

x+3 = 0

x+3 = 0 // - 3

x = -3

x in (-oo:-3) U (-3:4) U (4:+oo)

7/(x^2-x-12)+4/(x+3) = 1/(x-4) // - 1/(x-4)

7/(x^2-x-12)+4/(x+3)-(1/(x-4)) = 0

7/(x^2-x-12)+4/(x+3)-(x-4)^-1 = 0

7/(x^2-x-12)+4/(x+3)-1/(x-4) = 0

x^2-x-12 = 0

x^2-x-12 = 0

x^2-x-12 = 0

DELTA = (-1)^2-(-12*1*4)

DELTA = 49

DELTA > 0

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

x = 4 or x = -3

(x+3)*(x-4) = 0

7/((x+3)*(x-4))+4/(x+3)-1/(x-4) = 0

7/((x+3)*(x-4))+(4*(x-4))/((x+3)*(x-4))+(-1*(x+3))/((x+3)*(x-4)) = 0

4*(x-4)-1*(x+3)+7 = 0

4*x-x-9-3 = 0

3*x-12 = 0

(3*x-12)/((x+3)*(x-4)) = 0

(3*x-12)/((x+3)*(x-4)) = 0 // * (x+3)*(x-4)

3*x-12 = 0

3*x-12 = 0 // + 12

3*x = 12 // : 3

x = 12/3

x = 4

x in { 4} 

x belongs to the empty set

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