b/3=(b-9)/(2b-10)-(1/2)

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Solution for b/3=(b-9)/(2b-10)-(1/2) equation:


D( b )

2*b-10 = 0

2*b-10 = 0

2*b-10 = 0

2*b-10 = 0 // + 10

2*b = 10 // : 2

b = 10/2

b = 5

b in (-oo:5) U (5:+oo)

b/3 = (b-9)/(2*b-10)-(1/2) // - (b-9)/(2*b-10)-(1/2)

b/3-((b-9)/(2*b-10))+1/2 = 0

(-1*(b-9))/(2*b-10)+b/3+1/2 = 0

(-1*2*3*(b-9))/(2*3*(2*b-10))+(2*b*(2*b-10))/(2*3*(2*b-10))+(1*3*(2*b-10))/(2*3*(2*b-10)) = 0

2*b*(2*b-10)-1*2*3*(b-9)+1*3*(2*b-10) = 0

4*b^2-26*b+6*b-30+54 = 0

4*b^2-20*b+24 = 0

4*b^2-20*b+24 = 0

4*(b^2-5*b+6) = 0

b^2-5*b+6 = 0

DELTA = (-5)^2-(1*4*6)

DELTA = 1

DELTA > 0

b = (1^(1/2)+5)/(1*2) or b = (5-1^(1/2))/(1*2)

b = 3 or b = 2

4*(b-2)*(b-3) = 0

(4*(b-2)*(b-3))/(2*3*(2*b-10)) = 0

(4*(b-2)*(b-3))/(2*3*(2*b-10)) = 0 // * 2*3*(2*b-10)

4*(b-2)*(b-3) = 0

( 4 )

4 = 0

b belongs to the empty set

( b-2 )

b-2 = 0 // + 2

b = 2

( b-3 )

b-3 = 0 // + 3

b = 3

b in { 2, 3 }

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