1/3y+y=12

Solution for 1/3y+y=12 equation:

1/3y+y=12

We move all terms to the left:

1/3y+y-(12)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R


We add all the numbers together, and all the variables

y+1/3y-12=0

We multiply all the terms by the denominator


y*3y-12*3y+1=0

Wy multiply elements

3y^2-36y+1=0

a = 3; b = -36; c = +1;
Δ = b2-4ac
Δ = -362-4·3·1
Δ = 1284
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:

$\sqrt{\Delta}=\sqrt{1284}=\sqrt{4*321}=\sqrt{4}*\sqrt{321}=2\sqrt{321}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-2\sqrt{321}}{2*3}=\frac{36-2\sqrt{321}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+2\sqrt{321}}{2*3}=\frac{36+2\sqrt{321}}{6}
$