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