11b+1=(2/3)b+8

Solution for 11b+1=(2/3)b+8 equation:

11b+1=(2/3)b+8

We move all terms to the left:

11b+1-((2/3)b+8)=0


Domain of the equation: 3)b+8)!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

11b-((+2/3)b+8)+1=0

We multiply all the terms by the denominator


11b*3)b+8)-((+1*3)b+8)+2=0

We add all the numbers together, and all the variables

11b*3)b+8)-(3b+8)+2=0

Wy multiply elements

33b^2=0

a = 33; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·33·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$b=\frac{-b}{2a}=\frac{0}{66}=0$