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1/3b-1/2b=2+1
We move all terms to the left:
1/3b-1/2b-(2+1)=0
Domain of the equation: 3b!=0
b!=0/3
b!=0
b∈R
Domain of the equation: 2b!=0We add all the numbers together, and all the variables
b!=0/2
b!=0
b∈R
1/3b-1/2b-3=0
We calculate fractions
2b/6b^2+(-3b)/6b^2-3=0
We multiply all the terms by the denominator
2b+(-3b)-3*6b^2=0
Wy multiply elements
-18b^2+2b+(-3b)=0
We get rid of parentheses
-18b^2+2b-3b=0
We add all the numbers together, and all the variables
-18b^2-1b=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:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-18}=\frac{0}{-36} =0 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-18}=\frac{2}{-36} =-1/18 $
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