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2y+(7y-1/2)*2=2(1-5/2y)
We move all terms to the left:
2y+(7y-1/2)*2-(2(1-5/2y))=0
Domain of the equation: 2y))!=0We add all the numbers together, and all the variables
y!=0/1
y!=0
y∈R
2y+(+7y-1/2)*2-(2(-5/2y+1))=0
We multiply parentheses
2y+14y-(2(-5/2y+1))-1/2*2=0
We calculate fractions
2y+14y+()/2y+(-2y)/2y=0
We add all the numbers together, and all the variables
16y+()/2y+(-2y)/2y=0
We multiply all the terms by the denominator
16y*2y+(-2y)+()=0
We add all the numbers together, and all the variables
16y*2y+(-2y)=0
Wy multiply elements
32y^2+(-2y)=0
We get rid of parentheses
32y^2-2y=0
a = 32; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·32·0
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*32}=\frac{0}{64} =0 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*32}=\frac{4}{64} =1/16 $
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