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8/3x+5=3/x+3
We move all terms to the left:
8/3x+5-(3/x+3)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: x+3)!=0We get rid of parentheses
x∈R
8/3x-3/x-3+5=0
We calculate fractions
8x/3x^2+(-9x)/3x^2-3+5=0
We add all the numbers together, and all the variables
8x/3x^2+(-9x)/3x^2+2=0
We multiply all the terms by the denominator
8x+(-9x)+2*3x^2=0
Wy multiply elements
6x^2+8x+(-9x)=0
We get rid of parentheses
6x^2+8x-9x=0
We add all the numbers together, and all the variables
6x^2-1x=0
a = 6; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·6·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*6}=\frac{0}{12} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*6}=\frac{2}{12} =1/6 $
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