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4/3b=1/4b+3

We move all terms to the left:

4/3b-(1/4b+3)=0

Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R

Domain of the equation: 4b+3)!=0We get rid of parentheses

b∈R

4/3b-1/4b-3=0

We calculate fractions

16b/12b^2+(-3b)/12b^2-3=0

We multiply all the terms by the denominator

16b+(-3b)-3*12b^2=0

Wy multiply elements

-36b^2+16b+(-3b)=0

We get rid of parentheses

-36b^2+16b-3b=0

We add all the numbers together, and all the variables

-36b^2+13b=0

a = -36; b = 13; c = 0;

Δ = b^{2}-4ac

Δ = 13^{2}-4·(-36)·0

Δ = 169

The delta value is higher than zero, so the equation has two solutions

We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*-36}=\frac{-26}{-72} =13/36 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*-36}=\frac{0}{-72} =0 $

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