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3/4y+3=11+1/5y
We move all terms to the left:
3/4y+3-(11+1/5y)=0
Domain of the equation: 4y!=0
y!=0/4
y!=0
y∈R
Domain of the equation: 5y)!=0We add all the numbers together, and all the variables
y!=0/1
y!=0
y∈R
3/4y-(1/5y+11)+3=0
We get rid of parentheses
3/4y-1/5y-11+3=0
We calculate fractions
15y/20y^2+(-4y)/20y^2-11+3=0
We add all the numbers together, and all the variables
15y/20y^2+(-4y)/20y^2-8=0
We multiply all the terms by the denominator
15y+(-4y)-8*20y^2=0
Wy multiply elements
-160y^2+15y+(-4y)=0
We get rid of parentheses
-160y^2+15y-4y=0
We add all the numbers together, and all the variables
-160y^2+11y=0
a = -160; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·(-160)·0
Δ = 121
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{121}=11$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*-160}=\frac{-22}{-320} =11/160 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*-160}=\frac{0}{-320} =0 $
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