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x-5/3x+1=-1
We move all terms to the left:
x-5/3x+1-(-1)=0
Domain of the equation: 3x!=0We add all the numbers together, and all the variables
x!=0/3
x!=0
x∈R
x-5/3x+2=0
We multiply all the terms by the denominator
x*3x+2*3x-5=0
Wy multiply elements
3x^2+6x-5=0
a = 3; b = 6; c = -5;
Δ = b2-4ac
Δ = 62-4·3·(-5)
Δ = 96
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{6}}{2*3}=\frac{-6-4\sqrt{6}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{6}}{2*3}=\frac{-6+4\sqrt{6}}{6} $
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