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1/3x-3=-x+1
We move all terms to the left:
1/3x-3-(-x+1)=0
Domain of the equation: 3x!=0We add all the numbers together, and all the variables
x!=0/3
x!=0
x∈R
1/3x-(-1x+1)-3=0
We get rid of parentheses
1/3x+1x-1-3=0
We multiply all the terms by the denominator
1x*3x-1*3x-3*3x+1=0
Wy multiply elements
3x^2-3x-9x+1=0
We add all the numbers together, and all the variables
3x^2-12x+1=0
a = 3; b = -12; c = +1;
Δ = b2-4ac
Δ = -122-4·3·1
Δ = 132
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{33}}{2*3}=\frac{12-2\sqrt{33}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{33}}{2*3}=\frac{12+2\sqrt{33}}{6} $
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