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1/2y+3/4y=-20
We move all terms to the left:
1/2y+3/4y-(-20)=0
Domain of the equation: 2y!=0
y!=0/2
y!=0
y∈R
Domain of the equation: 4y!=0We add all the numbers together, and all the variables
y!=0/4
y!=0
y∈R
1/2y+3/4y+20=0
We calculate fractions
4y/8y^2+6y/8y^2+20=0
We multiply all the terms by the denominator
4y+6y+20*8y^2=0
We add all the numbers together, and all the variables
10y+20*8y^2=0
Wy multiply elements
160y^2+10y=0
a = 160; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·160·0
Δ = 100
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{100}=10$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*160}=\frac{-20}{320} =-1/16 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*160}=\frac{0}{320} =0 $
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