(1/3y)+(21/3)=3y-3

Solution for (1/3y)+(21/3)=3y-3 equation:

(1/3y)+(21/3)=3y-3

We move all terms to the left:

(1/3y)+(21/3)-(3y-3)=0


Domain of the equation: 3y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


-3y*3y+3*3y+7*3y+1=0

Wy multiply elements

-9y^2+9y+21y+1=0

We add all the numbers together, and all the variables

-9y^2+30y+1=0

a = -9; b = 30; c = +1;
Δ = b2-4ac
Δ = 302-4·(-9)·1
Δ = 936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{936}=\sqrt{36*26}=\sqrt{36}*\sqrt{26}=6\sqrt{26}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-6\sqrt{26}}{2*-9}=\frac{-30-6\sqrt{26}}{-18}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+6\sqrt{26}}{2*-9}=\frac{-30+6\sqrt{26}}{-18}
$