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1/2y-3=1/6y
We move all terms to the left:
1/2y-3-(1/6y)=0
Domain of the equation: 2y!=0
y!=0/2
y!=0
y∈R
Domain of the equation: 6y)!=0We add all the numbers together, and all the variables
y!=0/1
y!=0
y∈R
1/2y-(+1/6y)-3=0
We get rid of parentheses
1/2y-1/6y-3=0
We calculate fractions
6y/12y^2+(-2y)/12y^2-3=0
We multiply all the terms by the denominator
6y+(-2y)-3*12y^2=0
Wy multiply elements
-36y^2+6y+(-2y)=0
We get rid of parentheses
-36y^2+6y-2y=0
We add all the numbers together, and all the variables
-36y^2+4y=0
a = -36; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-36)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-36}=\frac{-8}{-72} =1/9 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-36}=\frac{0}{-72} =0 $
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