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1/3x+9=1/6x
We move all terms to the left:
1/3x+9-(1/6x)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 6x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
1/3x-(+1/6x)+9=0
We get rid of parentheses
1/3x-1/6x+9=0
We calculate fractions
6x/18x^2+(-3x)/18x^2+9=0
We multiply all the terms by the denominator
6x+(-3x)+9*18x^2=0
Wy multiply elements
162x^2+6x+(-3x)=0
We get rid of parentheses
162x^2+6x-3x=0
We add all the numbers together, and all the variables
162x^2+3x=0
a = 162; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·162·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*162}=\frac{-6}{324} =-1/54 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*162}=\frac{0}{324} =0 $
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