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1/2y+2=1/4y+5
We move all terms to the left:
1/2y+2-(1/4y+5)=0
Domain of the equation: 2y!=0
y!=0/2
y!=0
y∈R
Domain of the equation: 4y+5)!=0We get rid of parentheses
y∈R
1/2y-1/4y-5+2=0
We calculate fractions
4y/8y^2+(-2y)/8y^2-5+2=0
We add all the numbers together, and all the variables
4y/8y^2+(-2y)/8y^2-3=0
We multiply all the terms by the denominator
4y+(-2y)-3*8y^2=0
Wy multiply elements
-24y^2+4y+(-2y)=0
We get rid of parentheses
-24y^2+4y-2y=0
We add all the numbers together, and all the variables
-24y^2+2y=0
a = -24; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-24)·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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-24}=\frac{-4}{-48} =1/12 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-24}=\frac{0}{-48} =0 $
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