(1/3)x+(1/6)x=1/5

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

(1/3)x+(1/6)x=1/5

We move all terms to the left:

(1/3)x+(1/6)x-(1/5)=0


Domain of the equation: 3)x!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 6)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+1/3)x+(+1/6)x-(+1/5)=0

We multiply parentheses

x^2+x^2-(+1/5)=0

We get rid of parentheses

x^2+x^2-1/5=0

We multiply all the terms by the denominator


x^2*5+x^2*5-1=0

Wy multiply elements

5x^2+5x^2-1=0

We add all the numbers together, and all the variables

10x^2-1=0

a = 10; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·10·(-1)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{10}}{2*10}=\frac{0-2\sqrt{10}}{20}
=-\frac{2\sqrt{10}}{20}
=-\frac{\sqrt{10}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*10}=\frac{0+2\sqrt{10}}{20}
=\frac{2\sqrt{10}}{20}
=\frac{\sqrt{10}}{10}
$