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1/3x+2=1/6x+3
We move all terms to the left:
1/3x+2-(1/6x+3)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 6x+3)!=0We get rid of parentheses
x∈R
1/3x-1/6x-3+2=0
We calculate fractions
6x/18x^2+(-3x)/18x^2-3+2=0
We add all the numbers together, and all the variables
6x/18x^2+(-3x)/18x^2-1=0
We multiply all the terms by the denominator
6x+(-3x)-1*18x^2=0
Wy multiply elements
-18x^2+6x+(-3x)=0
We get rid of parentheses
-18x^2+6x-3x=0
We add all the numbers together, and all the variables
-18x^2+3x=0
a = -18; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-18)·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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-18}=\frac{-6}{-36} =1/6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-18}=\frac{0}{-36} =0 $
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