2/7y-3=3/9y-5

Solution for 2/7y-3=3/9y-5 equation:

2/7y-3=3/9y-5

We move all terms to the left:

2/7y-3-(3/9y-5)=0


Domain of the equation: 7y!=0

y!=0/7

y!=0

y∈R



Domain of the equation: 9y-5)!=0

y∈R


We get rid of parentheses

2/7y-3/9y+5-3=0

We calculate fractions


18y/63y^2+(-21y)/63y^2+5-3=0

We add all the numbers together, and all the variables

18y/63y^2+(-21y)/63y^2+2=0

We multiply all the terms by the denominator


18y+(-21y)+2*63y^2=0

Wy multiply elements

126y^2+18y+(-21y)=0

We get rid of parentheses

126y^2+18y-21y=0

We add all the numbers together, and all the variables

126y^2-3y=0

a = 126; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·126·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*126}=\frac{0}{252}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*126}=\frac{6}{252}
=1/42
$