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2/x+3=5/3x
We move all terms to the left:
2/x+3-(5/3x)=0
Domain of the equation: x!=0
x∈R
Domain of the equation: 3x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
2/x-(+5/3x)+3=0
We get rid of parentheses
2/x-5/3x+3=0
We calculate fractions
6x/3x^2+(-5x)/3x^2+3=0
We multiply all the terms by the denominator
6x+(-5x)+3*3x^2=0
Wy multiply elements
9x^2+6x+(-5x)=0
We get rid of parentheses
9x^2+6x-5x=0
We add all the numbers together, and all the variables
9x^2+x=0
a = 9; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·9·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*9}=\frac{-2}{18} =-1/9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*9}=\frac{0}{18} =0 $
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