2/y-3=4/3y+5

Solution for 2/y-3=4/3y+5 equation:

2/y-3=4/3y+5

We move all terms to the left:

2/y-3-(4/3y+5)=0


Domain of the equation: y!=0

y∈R



Domain of the equation: 3y+5)!=0

y∈R


We get rid of parentheses

2/y-4/3y-5-3=0

We calculate fractions


6y/3y^2+(-4y)/3y^2-5-3=0

We add all the numbers together, and all the variables

6y/3y^2+(-4y)/3y^2-8=0

We multiply all the terms by the denominator


6y+(-4y)-8*3y^2=0

Wy multiply elements

-24y^2+6y+(-4y)=0

We get rid of parentheses

-24y^2+6y-4y=0

We add all the numbers together, and all the variables

-24y^2+2y=0

a = -24; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-24)·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*-24}=\frac{-4}{-48}
=1/12
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-24}=\frac{0}{-48}
=0
$