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2(3x-1)-(2x+)/7x-2=1/3
We move all terms to the left:
2(3x-1)-(2x+)/7x-2-(1/3)=0
Domain of the equation: 7x!=0We add all the numbers together, and all the variables
x!=0/7
x!=0
x∈R
2(3x-1)-(+2x)/7x-2-(+1/3)=0
We multiply parentheses
6x-(+2x)/7x-2-2-(+1/3)=0
We get rid of parentheses
6x-(+2x)/7x-2-2-1/3=0
We calculate fractions
6x+(-6x)/21x+(-7x)/21x-2-2=0
We add all the numbers together, and all the variables
6x+(-6x)/21x+(-7x)/21x-4=0
We multiply all the terms by the denominator
6x*21x+(-6x)+(-7x)-4*21x=0
Wy multiply elements
126x^2+(-6x)+(-7x)-84x=0
We get rid of parentheses
126x^2-6x-7x-84x=0
We add all the numbers together, and all the variables
126x^2-97x=0
a = 126; b = -97; c = 0;
Δ = b2-4ac
Δ = -972-4·126·0
Δ = 9409
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{9409}=97$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-97)-97}{2*126}=\frac{0}{252} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-97)+97}{2*126}=\frac{194}{252} =97/126 $
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