(1/5)y+(2/3)=6

Solution for (1/5)y+(2/3)=6 equation:

(1/5)y+(2/3)=6

We move all terms to the left:

(1/5)y+(2/3)-(6)=0


Domain of the equation: 5)y!=0

y!=0/1

y!=0

y∈R


determiningTheFunctionDomain
(1/5)y-6+(2/3)=0

We add all the numbers together, and all the variables

(+1/5)y-6+(+2/3)=0

We multiply parentheses

y^2-6+(+2/3)=0

We get rid of parentheses

y^2-6+2/3=0

We multiply all the terms by the denominator


y^2*3+2-6*3=0

We add all the numbers together, and all the variables

y^2*3-16=0

Wy multiply elements

3y^2-16=0

a = 3; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·3·(-16)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*3}=\frac{0-8\sqrt{3}}{6}
=-\frac{8\sqrt{3}}{6}
=-\frac{4\sqrt{3}}{3}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*3}=\frac{0+8\sqrt{3}}{6}
=\frac{8\sqrt{3}}{6}
=\frac{4\sqrt{3}}{3}
$