1/5b+2-7=1/3b-3

Solution for 1/5b+2-7=1/3b-3 equation:

1/5b+2-7=1/3b-3

We move all terms to the left:

1/5b+2-7-(1/3b-3)=0


Domain of the equation: 5b!=0

b!=0/5

b!=0

b∈R



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

b∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

1/5b-1/3b+3-5=0

We calculate fractions


3b/15b^2+(-5b)/15b^2+3-5=0

We add all the numbers together, and all the variables

3b/15b^2+(-5b)/15b^2-2=0

We multiply all the terms by the denominator


3b+(-5b)-2*15b^2=0

Wy multiply elements

-30b^2+3b+(-5b)=0

We get rid of parentheses

-30b^2+3b-5b=0

We add all the numbers together, and all the variables

-30b^2-2b=0

a = -30; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-30)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-30}=\frac{0}{-60}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-30}=\frac{4}{-60}
=-1/15
$