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1/3y-10=1/9y
We move all terms to the left:
1/3y-10-(1/9y)=0
Domain of the equation: 3y!=0
y!=0/3
y!=0
y∈R
Domain of the equation: 9y)!=0We add all the numbers together, and all the variables
y!=0/1
y!=0
y∈R
1/3y-(+1/9y)-10=0
We get rid of parentheses
1/3y-1/9y-10=0
We calculate fractions
9y/27y^2+(-3y)/27y^2-10=0
We multiply all the terms by the denominator
9y+(-3y)-10*27y^2=0
Wy multiply elements
-270y^2+9y+(-3y)=0
We get rid of parentheses
-270y^2+9y-3y=0
We add all the numbers together, and all the variables
-270y^2+6y=0
a = -270; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-270)·0
Δ = 36
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{36}=6$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-270}=\frac{-12}{-540} =1/45 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-270}=\frac{0}{-540} =0 $
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