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1/6x+3=2/3x+9
We move all terms to the left:
1/6x+3-(2/3x+9)=0
Domain of the equation: 6x!=0
x!=0/6
x!=0
x∈R
Domain of the equation: 3x+9)!=0We get rid of parentheses
x∈R
1/6x-2/3x-9+3=0
We calculate fractions
3x/18x^2+(-12x)/18x^2-9+3=0
We add all the numbers together, and all the variables
3x/18x^2+(-12x)/18x^2-6=0
We multiply all the terms by the denominator
3x+(-12x)-6*18x^2=0
Wy multiply elements
-108x^2+3x+(-12x)=0
We get rid of parentheses
-108x^2+3x-12x=0
We add all the numbers together, and all the variables
-108x^2-9x=0
a = -108; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·(-108)·0
Δ = 81
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{81}=9$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*-108}=\frac{0}{-216} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*-108}=\frac{18}{-216} =-1/12 $
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