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x-5/3x+2=x+1/x
We move all terms to the left:
x-5/3x+2-(x+1/x)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
x-5/3x-(+x+1/x)+2=0
We get rid of parentheses
x-5/3x-x-1/x+2=0
We calculate fractions
x-x+(-5x)/3x^2+(-3x)/3x^2+2=0
We add all the numbers together, and all the variables
(-5x)/3x^2+(-3x)/3x^2+2=0
We multiply all the terms by the denominator
(-5x)+(-3x)+2*3x^2=0
Wy multiply elements
6x^2+(-5x)+(-3x)=0
We get rid of parentheses
6x^2-5x-3x=0
We add all the numbers together, and all the variables
6x^2-8x=0
a = 6; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·6·0
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*6}=\frac{0}{12} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*6}=\frac{16}{12} =1+1/3 $
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