1/3b+1=13-1/9b

Solution for 1/3b+1=13-1/9b equation:

1/3b+1=13-1/9b

We move all terms to the left:

1/3b+1-(13-1/9b)=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-(-1/9b+13)+1=0

We get rid of parentheses

1/3b+1/9b-13+1=0

We calculate fractions


9b/27b^2+3b/27b^2-13+1=0

We add all the numbers together, and all the variables

9b/27b^2+3b/27b^2-12=0

We multiply all the terms by the denominator


9b+3b-12*27b^2=0

We add all the numbers together, and all the variables

12b-12*27b^2=0

Wy multiply elements

-324b^2+12b=0

a = -324; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-324)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-324}=\frac{-24}{-648}
=1/27
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-324}=\frac{0}{-648}
=0
$