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