1/3b-10=2/9b

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

1/3b-10=2/9b

We move all terms to the left:

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


Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R



Domain of the equation: 9b)!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

1/3b-2/9b-10=0

We calculate fractions


9b/27b^2+(-6b)/27b^2-10=0

We multiply all the terms by the denominator


9b+(-6b)-10*27b^2=0

Wy multiply elements

-270b^2+9b+(-6b)=0

We get rid of parentheses

-270b^2+9b-6b=0

We add all the numbers together, and all the variables

-270b^2+3b=0

a = -270; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-270)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-270}=\frac{-6}{-540}
=1/90
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-270}=\frac{0}{-540}
=0
$