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1/3y-5=1/2y
We move all terms to the left:
1/3y-5-(1/2y)=0
Domain of the equation: 3y!=0
y!=0/3
y!=0
y∈R
Domain of the equation: 2y)!=0We add all the numbers together, and all the variables
y!=0/1
y!=0
y∈R
1/3y-(+1/2y)-5=0
We get rid of parentheses
1/3y-1/2y-5=0
We calculate fractions
2y/6y^2+(-3y)/6y^2-5=0
We multiply all the terms by the denominator
2y+(-3y)-5*6y^2=0
Wy multiply elements
-30y^2+2y+(-3y)=0
We get rid of parentheses
-30y^2+2y-3y=0
We add all the numbers together, and all the variables
-30y^2-1y=0
a = -30; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-30)·0
Δ = 1
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{1}=1$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-30}=\frac{0}{-60} =0 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-30}=\frac{2}{-60} =-1/30 $
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