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2/5y-3=1/2y
We move all terms to the left:
2/5y-3-(1/2y)=0
Domain of the equation: 5y!=0
y!=0/5
y!=0
y∈R
Domain of the equation: 2y)!=0We add all the numbers together, and all the variables
y!=0/1
y!=0
y∈R
2/5y-(+1/2y)-3=0
We get rid of parentheses
2/5y-1/2y-3=0
We calculate fractions
4y/10y^2+(-5y)/10y^2-3=0
We multiply all the terms by the denominator
4y+(-5y)-3*10y^2=0
Wy multiply elements
-30y^2+4y+(-5y)=0
We get rid of parentheses
-30y^2+4y-5y=0
We add all the numbers together, and all the variables
-30y^2-1y=0
a = -30; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-30)·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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-30}=\frac{0}{-60} =0 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-30}=\frac{2}{-60} =-1/30 $
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