1/3y+8=1/10y

Solution for 1/3y+8=1/10y equation:

1/3y+8=1/10y

We move all terms to the left:

1/3y+8-(1/10y)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



Domain of the equation: 10y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

1/3y-(+1/10y)+8=0

We get rid of parentheses

1/3y-1/10y+8=0

We calculate fractions


10y/30y^2+(-3y)/30y^2+8=0

We multiply all the terms by the denominator


10y+(-3y)+8*30y^2=0

Wy multiply elements

240y^2+10y+(-3y)=0

We get rid of parentheses

240y^2+10y-3y=0

We add all the numbers together, and all the variables

240y^2+7y=0

a = 240; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·240·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*240}=\frac{-14}{480}
=-7/240
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*240}=\frac{0}{480}
=0
$