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x-4/3x+3=2
We move all terms to the left:
x-4/3x+3-(2)=0
Domain of the equation: 3x!=0We add all the numbers together, and all the variables
x!=0/3
x!=0
x∈R
x-4/3x+1=0
We multiply all the terms by the denominator
x*3x+1*3x-4=0
Wy multiply elements
3x^2+3x-4=0
a = 3; b = 3; c = -4;
Δ = b2-4ac
Δ = 32-4·3·(-4)
Δ = 57
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{57}}{2*3}=\frac{-3-\sqrt{57}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{57}}{2*3}=\frac{-3+\sqrt{57}}{6} $
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