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1/3y-5=1/6y
We move all terms to the left:
1/3y-5-(1/6y)=0
Domain of the equation: 3y!=0
y!=0/3
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
1/3y-(+1/6y)-5=0
We get rid of parentheses
1/3y-1/6y-5=0
We calculate fractions
6y/18y^2+(-3y)/18y^2-5=0
We multiply all the terms by the denominator
6y+(-3y)-5*18y^2=0
Wy multiply elements
-90y^2+6y+(-3y)=0
We get rid of parentheses
-90y^2+6y-3y=0
We add all the numbers together, and all the variables
-90y^2+3y=0
a = -90; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-90)·0
Δ = 9
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{9}=3$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-90}=\frac{-6}{-180} =1/30 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-90}=\frac{0}{-180} =0 $
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