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1/2y+3=1/8y
We move all terms to the left:
1/2y+3-(1/8y)=0
Domain of the equation: 2y!=0
y!=0/2
y!=0
y∈R
Domain of the equation: 8y)!=0We add all the numbers together, and all the variables
y!=0/1
y!=0
y∈R
1/2y-(+1/8y)+3=0
We get rid of parentheses
1/2y-1/8y+3=0
We calculate fractions
8y/16y^2+(-2y)/16y^2+3=0
We multiply all the terms by the denominator
8y+(-2y)+3*16y^2=0
Wy multiply elements
48y^2+8y+(-2y)=0
We get rid of parentheses
48y^2+8y-2y=0
We add all the numbers together, and all the variables
48y^2+6y=0
a = 48; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·48·0
Δ = 36
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{36}=6$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*48}=\frac{-12}{96} =-1/8 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*48}=\frac{0}{96} =0 $
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