8/(3k-9)-5/(k-3)=4

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Solution for 8/(3k-9)-5/(k-3)=4 equation: 


D( k )

k-3 = 0

3*k-9 = 0

k-3 = 0

k-3 = 0

k-3 = 0 // + 3

k = 3

3*k-9 = 0

3*k-9 = 0

3*k-9 = 0 // + 9

3*k = 9 // : 3

k = 9/3

k = 3

k in (-oo:3) U (3:+oo)

8/(3*k-9)-(5/(k-3)) = 4 // - 4

8/(3*k-9)-(5/(k-3))-4 = 0

8/(3*k-9)-5*(k-3)^-1-4 = 0

8/(3*k-9)-5/(k-3)-4 = 0

(8*(k-3))/((3*k-9)*(k-3))+(-5*(3*k-9))/((3*k-9)*(k-3))+(-4*(3*k-9)*(k-3))/((3*k-9)*(k-3)) = 0

8*(k-3)-5*(3*k-9)-4*(3*k-9)*(k-3) = 0

72*k-12*k^2-7*k-108+21 = 0

65*k-12*k^2-87 = 0

65*k-12*k^2-87 = 0

65*k-12*k^2-87 = 0

DELTA = 65^2-(-87*(-12)*4)

DELTA = 49

DELTA > 0

k = (49^(1/2)-65)/(-12*2) or k = (-49^(1/2)-65)/(-12*2)

k = 29/12 or k = 3

(k-29/12)*(k-3) = 0

((k-29/12)*(k-3))/((3*k-9)*(k-3)) = 0

((k-29/12)*(k-3))/((3*k-9)*(k-3)) = 0 // * (3*k-9)*(k-3)

(k-29/12)*(k-3) = 0

( k-29/12 )

k-29/12 = 0 // + 29/12

k = 29/12

( k-3 )

k-3 = 0 // + 3

k = 3

k in { 3} 

k = 29/12

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