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1/6x+2=1/3x-3=
We move all terms to the left:
1/6x+2-(1/3x-3)=0
Domain of the equation: 6x!=0
x!=0/6
x!=0
x∈R
Domain of the equation: 3x-3)!=0We get rid of parentheses
x∈R
1/6x-1/3x+3+2=0
We calculate fractions
3x/18x^2+(-6x)/18x^2+3+2=0
We add all the numbers together, and all the variables
3x/18x^2+(-6x)/18x^2+5=0
We multiply all the terms by the denominator
3x+(-6x)+5*18x^2=0
Wy multiply elements
90x^2+3x+(-6x)=0
We get rid of parentheses
90x^2+3x-6x=0
We add all the numbers together, and all the variables
90x^2-3x=0
a = 90; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·90·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*90}=\frac{0}{180} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*90}=\frac{6}{180} =1/30 $
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