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2x+1/6=6(1/3x+1)
We move all terms to the left:
2x+1/6-(6(1/3x+1))=0
Domain of the equation: 3x+1))!=0We calculate fractions
x∈R
2x+()/3x+3x/3x=0
Fractions to decimals
()/3x+2x+1=0
We multiply all the terms by the denominator
2x*3x+1*3x+()=0
We add all the numbers together, and all the variables
2x*3x+1*3x=0
Wy multiply elements
6x^2+3x=0
a = 6; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·6·0
Δ = 9
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*6}=\frac{-6}{12} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*6}=\frac{0}{12} =0 $
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