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5/3x+1/3x=3x
We move all terms to the left:
5/3x+1/3x-(3x)=0
Domain of the equation: 3x!=0We add all the numbers together, and all the variables
x!=0/3
x!=0
x∈R
-3x+5/3x+1/3x=0
We multiply all the terms by the denominator
-3x*3x+5+1=0
We add all the numbers together, and all the variables
-3x*3x+6=0
Wy multiply elements
-9x^2+6=0
a = -9; b = 0; c = +6;
Δ = b2-4ac
Δ = 02-4·(-9)·6
Δ = 216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{6}}{2*-9}=\frac{0-6\sqrt{6}}{-18} =-\frac{6\sqrt{6}}{-18} =-\frac{\sqrt{6}}{-3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{6}}{2*-9}=\frac{0+6\sqrt{6}}{-18} =\frac{6\sqrt{6}}{-18} =\frac{\sqrt{6}}{-3} $
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