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2/3a=3+1/2a
We move all terms to the left:
2/3a-(3+1/2a)=0
Domain of the equation: 3a!=0
a!=0/3
a!=0
a∈R
Domain of the equation: 2a)!=0We add all the numbers together, and all the variables
a!=0/1
a!=0
a∈R
2/3a-(1/2a+3)=0
We get rid of parentheses
2/3a-1/2a-3=0
We calculate fractions
4a/6a^2+(-3a)/6a^2-3=0
We multiply all the terms by the denominator
4a+(-3a)-3*6a^2=0
Wy multiply elements
-18a^2+4a+(-3a)=0
We get rid of parentheses
-18a^2+4a-3a=0
We add all the numbers together, and all the variables
-18a^2+a=0
a = -18; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-18)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-18}=\frac{-2}{-36} =1/18 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-18}=\frac{0}{-36} =0 $
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