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

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

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

We move all terms to the left:

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


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

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

Wy multiply elements

9y^2-6y=0

a = 9; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·9·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*9}=\frac{0}{18}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*9}=\frac{12}{18}
=2/3
$