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(1/2x+1)-(1/3x-1)=0
Domain of the equation: 2x+1)!=0
x∈R
Domain of the equation: 3x-1)!=0We get rid of parentheses
x∈R
1/2x-1/3x+1+1=0
We calculate fractions
3x/6x^2+(-2x)/6x^2+1+1=0
We add all the numbers together, and all the variables
3x/6x^2+(-2x)/6x^2+2=0
We multiply all the terms by the denominator
3x+(-2x)+2*6x^2=0
Wy multiply elements
12x^2+3x+(-2x)=0
We get rid of parentheses
12x^2+3x-2x=0
We add all the numbers together, and all the variables
12x^2+x=0
a = 12; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·12·0
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*12}=\frac{-2}{24} =-1/12 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*12}=\frac{0}{24} =0 $
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