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1/3x+3=x
We move all terms to the left:
1/3x+3-(x)=0
Domain of the equation: 3x!=0We add all the numbers together, and all the variables
x!=0/3
x!=0
x∈R
-1x+1/3x+3=0
We multiply all the terms by the denominator
-1x*3x+3*3x+1=0
Wy multiply elements
-3x^2+9x+1=0
a = -3; b = 9; c = +1;
Δ = b2-4ac
Δ = 92-4·(-3)·1
Δ = 93
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{93}}{2*-3}=\frac{-9-\sqrt{93}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{93}}{2*-3}=\frac{-9+\sqrt{93}}{-6} $
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