(1)/(t+4)+(9)/(t+5)=(1)/(t^2+9t+20)

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Solution for (1)/(t+4)+(9)/(t+5)=(1)/(t^2+9t+20) equation:


D( t )

t^2+9*t+20 = 0

t+5 = 0

t+4 = 0

t^2+9*t+20 = 0

t^2+9*t+20 = 0

t^2+9*t+20 = 0

DELTA = 9^2-(1*4*20)

DELTA = 1

DELTA > 0

t = (1^(1/2)-9)/(1*2) or t = (-1^(1/2)-9)/(1*2)

t = -4 or t = -5

t+5 = 0

t+5 = 0

t+5 = 0 // - 5

t = -5

t+4 = 0

t+4 = 0

t+4 = 0 // - 4

t = -4

t in (-oo:-5) U (-5:-4) U (-4:+oo)

1/(t+4)+9/(t+5) = 1/(t^2+9*t+20) // - 1/(t^2+9*t+20)

1/(t+4)+9/(t+5)-(1/(t^2+9*t+20)) = 0

1/(t+4)+9/(t+5)-(t^2+9*t+20)^-1 = 0

1/(t+4)+9/(t+5)-1/(t^2+9*t+20) = 0

t^2+9*t+20 = 0

t^2+9*t+20 = 0

t^2+9*t+20 = 0

DELTA = 9^2-(1*4*20)

DELTA = 1

DELTA > 0

t = (1^(1/2)-9)/(1*2) or t = (-1^(1/2)-9)/(1*2)

t = -4 or t = -5

(t+5)*(t+4) = 0

1/(t+4)+9/(t+5)-1/((t+5)*(t+4)) = 0

(1*(t+5))/((t+4)*(t+5))+(9*(t+4))/((t+4)*(t+5))-1/((t+4)*(t+5)) = 0

1*(t+5)+9*(t+4)-1 = 0

10*t-1+41 = 0

10*t+40 = 0

(10*t+40)/((t+4)*(t+5)) = 0

(10*t+40)/((t+4)*(t+5)) = 0 // * (t+4)*(t+5)

10*t+40 = 0

10*t+40 = 0 // - 40

10*t = -40 // : 10

t = -40/10

t = -4

t in { -4}

t belongs to the empty set

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