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1/6g+3=4-1/3g
We move all terms to the left:
1/6g+3-(4-1/3g)=0
Domain of the equation: 6g!=0
g!=0/6
g!=0
g∈R
Domain of the equation: 3g)!=0We add all the numbers together, and all the variables
g!=0/1
g!=0
g∈R
1/6g-(-1/3g+4)+3=0
We get rid of parentheses
1/6g+1/3g-4+3=0
We calculate fractions
3g/18g^2+6g/18g^2-4+3=0
We add all the numbers together, and all the variables
3g/18g^2+6g/18g^2-1=0
We multiply all the terms by the denominator
3g+6g-1*18g^2=0
We add all the numbers together, and all the variables
9g-1*18g^2=0
Wy multiply elements
-18g^2+9g=0
a = -18; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·(-18)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{81}=9$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*-18}=\frac{-18}{-36} =1/2 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*-18}=\frac{0}{-36} =0 $
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