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1/4y+5=1/8y
We move all terms to the left:
1/4y+5-(1/8y)=0
Domain of the equation: 4y!=0
y!=0/4
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
1/4y-(+1/8y)+5=0
We get rid of parentheses
1/4y-1/8y+5=0
We calculate fractions
8y/32y^2+(-4y)/32y^2+5=0
We multiply all the terms by the denominator
8y+(-4y)+5*32y^2=0
Wy multiply elements
160y^2+8y+(-4y)=0
We get rid of parentheses
160y^2+8y-4y=0
We add all the numbers together, and all the variables
160y^2+4y=0
a = 160; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·160·0
Δ = 16
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{16}=4$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*160}=\frac{-8}{320} =-1/40 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*160}=\frac{0}{320} =0 $
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