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1/3x-3=1/9x=3
We move all terms to the left:
1/3x-3-(1/9x)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 9x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
1/3x-(+1/9x)-3=0
We get rid of parentheses
1/3x-1/9x-3=0
We calculate fractions
9x/27x^2+(-3x)/27x^2-3=0
We multiply all the terms by the denominator
9x+(-3x)-3*27x^2=0
Wy multiply elements
-81x^2+9x+(-3x)=0
We get rid of parentheses
-81x^2+9x-3x=0
We add all the numbers together, and all the variables
-81x^2+6x=0
a = -81; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-81)·0
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-81}=\frac{-12}{-162} =2/27 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-81}=\frac{0}{-162} =0 $
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