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1/2y-1=1/5y
We move all terms to the left:
1/2y-1-(1/5y)=0
Domain of the equation: 2y!=0
y!=0/2
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
1/2y-(+1/5y)-1=0
We get rid of parentheses
1/2y-1/5y-1=0
We calculate fractions
5y/10y^2+(-2y)/10y^2-1=0
We multiply all the terms by the denominator
5y+(-2y)-1*10y^2=0
Wy multiply elements
-10y^2+5y+(-2y)=0
We get rid of parentheses
-10y^2+5y-2y=0
We add all the numbers together, and all the variables
-10y^2+3y=0
a = -10; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-10)·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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-10}=\frac{-6}{-20} =3/10 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-10}=\frac{0}{-20} =0 $
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