(5/3+b)=(3/7b+1)

Solution for (5/3+b)=(3/7b+1) equation:

(5/3+b)=(3/7b+1)

We move all terms to the left:

(5/3+b)-((3/7b+1))=0


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

We move all terms containing b to the left, all other terms to the right

b)!=-3

b!=-3/1

b!=-3

b∈R



Domain of the equation: 7b+1))!=0

b∈R


We add all the numbers together, and all the variables

(+b+5/3)-((3/7b+1))=0

We get rid of parentheses

b-((3/7b+1))+5/3=0

We calculate fractions


b+()/7b+35b/7b=0

We multiply all the terms by the denominator


b*7b+35b+()=0

We add all the numbers together, and all the variables

35b+b*7b=0

Wy multiply elements

7b^2+35b=0

a = 7; b = 35; c = 0;
Δ = b2-4ac
Δ = 352-4·7·0
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-35}{2*7}=\frac{-70}{14}
=-5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+35}{2*7}=\frac{0}{14}
=0
$