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

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

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

We move all terms to the left:

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


Domain of the equation: 3)(6x-3)!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply parentheses ..

(+6x^2+1/3*-3)-x-5=0

We multiply all the terms by the denominator


(+6x^2+1-x*3*-3)-5*3*-3)=0

We add all the numbers together, and all the variables

(+6x^2+1-x*3*-3)=0

We get rid of parentheses

6x^2-x*3*+1-3=0

We add all the numbers together, and all the variables

6x^2-x*3*-2=0

Wy multiply elements

6x^2-3x^2-2=0

We add all the numbers together, and all the variables

3x^2-2=0

a = 3; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·3·(-2)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*3}=\frac{0-2\sqrt{6}}{6}
=-\frac{2\sqrt{6}}{6}
=-\frac{\sqrt{6}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*3}=\frac{0+2\sqrt{6}}{6}
=\frac{2\sqrt{6}}{6}
=\frac{\sqrt{6}}{3}
$