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1/3y+8=1/5y
We move all terms to the left:
1/3y+8-(1/5y)=0
Domain of the equation: 3y!=0
y!=0/3
y!=0
y∈R
Domain of the equation: 5y)!=0We add all the numbers together, and all the variables
y!=0/1
y!=0
y∈R
1/3y-(+1/5y)+8=0
We get rid of parentheses
1/3y-1/5y+8=0
We calculate fractions
5y/15y^2+(-3y)/15y^2+8=0
We multiply all the terms by the denominator
5y+(-3y)+8*15y^2=0
Wy multiply elements
120y^2+5y+(-3y)=0
We get rid of parentheses
120y^2+5y-3y=0
We add all the numbers together, and all the variables
120y^2+2y=0
a = 120; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·120·0
Δ = 4
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{4}=2$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*120}=\frac{-4}{240} =-1/60 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*120}=\frac{0}{240} =0 $
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