(1/6)y+(2/12)=2

Solution for (1/6)y+(2/12)=2 equation:

(1/6)y+(2/12)=2

We move all terms to the left:

(1/6)y+(2/12)-(2)=0


Domain of the equation: 6)y!=0

y!=0/1

y!=0

y∈R


determiningTheFunctionDomain
(1/6)y-2+(2/12)=0

We add all the numbers together, and all the variables

(+1/6)y-2+(+2/12)=0

We multiply parentheses

y^2-2+(+2/12)=0

We get rid of parentheses

y^2-2+2/12=0

We multiply all the terms by the denominator


y^2*12+2-2*12=0

We add all the numbers together, and all the variables

y^2*12-22=0

Wy multiply elements

12y^2-22=0

a = 12; b = 0; c = -22;
Δ = b2-4ac
Δ = 02-4·12·(-22)
Δ = 1056
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:

$\sqrt{\Delta}=\sqrt{1056}=\sqrt{16*66}=\sqrt{16}*\sqrt{66}=4\sqrt{66}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{66}}{2*12}=\frac{0-4\sqrt{66}}{24}
=-\frac{4\sqrt{66}}{24}
=-\frac{\sqrt{66}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{66}}{2*12}=\frac{0+4\sqrt{66}}{24}
=\frac{4\sqrt{66}}{24}
=\frac{\sqrt{66}}{6}
$