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