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1/3x+4=-3x+1
We move all terms to the left:
1/3x+4-(-3x+1)=0
Domain of the equation: 3x!=0We get rid of parentheses
x!=0/3
x!=0
x∈R
1/3x+3x-1+4=0
We multiply all the terms by the denominator
3x*3x-1*3x+4*3x+1=0
Wy multiply elements
9x^2-3x+12x+1=0
We add all the numbers together, and all the variables
9x^2+9x+1=0
a = 9; b = 9; c = +1;
Δ = b2-4ac
Δ = 92-4·9·1
Δ = 45
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{5}}{2*9}=\frac{-9-3\sqrt{5}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{5}}{2*9}=\frac{-9+3\sqrt{5}}{18} $
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