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3/4y-3=1/2y+2
We move all terms to the left:
3/4y-3-(1/2y+2)=0
Domain of the equation: 4y!=0
y!=0/4
y!=0
y∈R
Domain of the equation: 2y+2)!=0We get rid of parentheses
y∈R
3/4y-1/2y-2-3=0
We calculate fractions
6y/8y^2+(-4y)/8y^2-2-3=0
We add all the numbers together, and all the variables
6y/8y^2+(-4y)/8y^2-5=0
We multiply all the terms by the denominator
6y+(-4y)-5*8y^2=0
Wy multiply elements
-40y^2+6y+(-4y)=0
We get rid of parentheses
-40y^2+6y-4y=0
We add all the numbers together, and all the variables
-40y^2+2y=0
a = -40; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-40)·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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-40}=\frac{-4}{-80} =1/20 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-40}=\frac{0}{-80} =0 $
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