(2/3y)+y-4=31

Solution for (2/3y)+y-4=31 equation:

(2/3y)+y-4=31

We move all terms to the left:

(2/3y)+y-4-(31)=0


Domain of the equation: 3y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

(+2/3y)+y-4-31=0

We add all the numbers together, and all the variables

y+(+2/3y)-35=0

We get rid of parentheses

y+2/3y-35=0

We multiply all the terms by the denominator


y*3y-35*3y+2=0

Wy multiply elements

3y^2-105y+2=0

a = 3; b = -105; c = +2;
Δ = b2-4ac
Δ = -1052-4·3·2
Δ = 11001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-105)-\sqrt{11001}}{2*3}=\frac{105-\sqrt{11001}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-105)+\sqrt{11001}}{2*3}=\frac{105+\sqrt{11001}}{6}
$