3+2/3b=11-2/5b

Solution for 3+2/3b=11-2/5b equation:

3+2/3b=11-2/5b

We move all terms to the left:

3+2/3b-(11-2/5b)=0


Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R



Domain of the equation: 5b)!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

2/3b-(-2/5b+11)+3=0

We get rid of parentheses

2/3b+2/5b-11+3=0

We calculate fractions


10b/15b^2+6b/15b^2-11+3=0

We add all the numbers together, and all the variables

10b/15b^2+6b/15b^2-8=0

We multiply all the terms by the denominator


10b+6b-8*15b^2=0

We add all the numbers together, and all the variables

16b-8*15b^2=0

Wy multiply elements

-120b^2+16b=0

a = -120; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-120)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-120}=\frac{-32}{-240}
=2/15
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-120}=\frac{0}{-240}
=0
$