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1/2x-3=1/6x+5
We move all terms to the left:
1/2x-3-(1/6x+5)=0
Domain of the equation: 2x!=0
x!=0/2
x!=0
x∈R
Domain of the equation: 6x+5)!=0We get rid of parentheses
x∈R
1/2x-1/6x-5-3=0
We calculate fractions
6x/12x^2+(-2x)/12x^2-5-3=0
We add all the numbers together, and all the variables
6x/12x^2+(-2x)/12x^2-8=0
We multiply all the terms by the denominator
6x+(-2x)-8*12x^2=0
Wy multiply elements
-96x^2+6x+(-2x)=0
We get rid of parentheses
-96x^2+6x-2x=0
We add all the numbers together, and all the variables
-96x^2+4x=0
a = -96; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-96)·0
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-96}=\frac{-8}{-192} =1/24 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-96}=\frac{0}{-192} =0 $
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