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1/3x-2/9x=10
We move all terms to the left:
1/3x-2/9x-(10)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 9x!=0We calculate fractions
x!=0/9
x!=0
x∈R
9x/27x^2+(-6x)/27x^2-10=0
We multiply all the terms by the denominator
9x+(-6x)-10*27x^2=0
Wy multiply elements
-270x^2+9x+(-6x)=0
We get rid of parentheses
-270x^2+9x-6x=0
We add all the numbers together, and all the variables
-270x^2+3x=0
a = -270; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-270)·0
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-270}=\frac{-6}{-540} =1/90 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-270}=\frac{0}{-540} =0 $
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