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1/4y+2=1/5y
We move all terms to the left:
1/4y+2-(1/5y)=0
Domain of the equation: 4y!=0
y!=0/4
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
1/4y-(+1/5y)+2=0
We get rid of parentheses
1/4y-1/5y+2=0
We calculate fractions
5y/20y^2+(-4y)/20y^2+2=0
We multiply all the terms by the denominator
5y+(-4y)+2*20y^2=0
Wy multiply elements
40y^2+5y+(-4y)=0
We get rid of parentheses
40y^2+5y-4y=0
We add all the numbers together, and all the variables
40y^2+y=0
a = 40; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·40·0
Δ = 1
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{1}=1$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*40}=\frac{-2}{80} =-1/40 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*40}=\frac{0}{80} =0 $
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