1/3y+7=1/5y

Solution for 1/3y+7=1/5y equation:

1/3y+7=1/5y

We move all terms to the left:

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


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



Domain of the equation: 5y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

1/3y-1/5y+7=0

We calculate fractions


5y/15y^2+(-3y)/15y^2+7=0

We multiply all the terms by the denominator


5y+(-3y)+7*15y^2=0

Wy multiply elements

105y^2+5y+(-3y)=0

We get rid of parentheses

105y^2+5y-3y=0

We add all the numbers together, and all the variables

105y^2+2y=0

a = 105; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·105·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*105}=\frac{-4}{210}
=-2/105
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*105}=\frac{0}{210}
=0
$