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1/2y+9=1/4y+4
We move all terms to the left:
1/2y+9-(1/4y+4)=0
Domain of the equation: 2y!=0
y!=0/2
y!=0
y∈R
Domain of the equation: 4y+4)!=0We get rid of parentheses
y∈R
1/2y-1/4y-4+9=0
We calculate fractions
4y/8y^2+(-2y)/8y^2-4+9=0
We add all the numbers together, and all the variables
4y/8y^2+(-2y)/8y^2+5=0
We multiply all the terms by the denominator
4y+(-2y)+5*8y^2=0
Wy multiply elements
40y^2+4y+(-2y)=0
We get rid of parentheses
40y^2+4y-2y=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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