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3/2y+12=7/5y-20
We move all terms to the left:
3/2y+12-(7/5y-20)=0
Domain of the equation: 2y!=0
y!=0/2
y!=0
y∈R
Domain of the equation: 5y-20)!=0We get rid of parentheses
y∈R
3/2y-7/5y+20+12=0
We calculate fractions
15y/10y^2+(-14y)/10y^2+20+12=0
We add all the numbers together, and all the variables
15y/10y^2+(-14y)/10y^2+32=0
We multiply all the terms by the denominator
15y+(-14y)+32*10y^2=0
Wy multiply elements
320y^2+15y+(-14y)=0
We get rid of parentheses
320y^2+15y-14y=0
We add all the numbers together, and all the variables
320y^2+y=0
a = 320; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·320·0
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*320}=\frac{-2}{640} =-1/320 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*320}=\frac{0}{640} =0 $
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