(6/9b)-(5/3b)=9

Solution for (6/9b)-(5/3b)=9 equation:

(6/9b)-(5/3b)=9

We move all terms to the left:

(6/9b)-(5/3b)-(9)=0


Domain of the equation: 9b)!=0

b!=0/1

b!=0

b∈R



Domain of the equation: 3b)!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

(+6/9b)-(+5/3b)-9=0

We get rid of parentheses

6/9b-5/3b-9=0

We calculate fractions


18b/27b^2+(-45b)/27b^2-9=0

We multiply all the terms by the denominator


18b+(-45b)-9*27b^2=0

Wy multiply elements

-243b^2+18b+(-45b)=0

We get rid of parentheses

-243b^2+18b-45b=0

We add all the numbers together, and all the variables

-243b^2-27b=0

a = -243; b = -27; c = 0;
Δ = b2-4ac
Δ = -272-4·(-243)·0
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-27}{2*-243}=\frac{0}{-486}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+27}{2*-243}=\frac{54}{-486}
=-1/9
$