(1/5)x+(1/3)=3

Solution for (1/5)x+(1/3)=3 equation:

(1/5)x+(1/3)=3

We move all terms to the left:

(1/5)x+(1/3)-(3)=0


Domain of the equation: 5)x!=0

x!=0/1

x!=0

x∈R


determiningTheFunctionDomain
(1/5)x-3+(1/3)=0

We add all the numbers together, and all the variables

(+1/5)x-3+(+1/3)=0

We multiply parentheses

x^2-3+(+1/3)=0

We get rid of parentheses

x^2-3+1/3=0

We multiply all the terms by the denominator


x^2*3+1-3*3=0

We add all the numbers together, and all the variables

x^2*3-8=0

Wy multiply elements

3x^2-8=0

a = 3; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·3·(-8)
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*3}=\frac{0-4\sqrt{6}}{6}
=-\frac{4\sqrt{6}}{6}
=-\frac{2\sqrt{6}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*3}=\frac{0+4\sqrt{6}}{6}
=\frac{4\sqrt{6}}{6}
=\frac{2\sqrt{6}}{3}
$