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7/3y=2/y+4
We move all terms to the left:
7/3y-(2/y+4)=0
Domain of the equation: 3y!=0
y!=0/3
y!=0
y∈R
Domain of the equation: y+4)!=0We get rid of parentheses
y∈R
7/3y-2/y-4=0
We calculate fractions
7y/3y^2+(-6y)/3y^2-4=0
We multiply all the terms by the denominator
7y+(-6y)-4*3y^2=0
Wy multiply elements
-12y^2+7y+(-6y)=0
We get rid of parentheses
-12y^2+7y-6y=0
We add all the numbers together, and all the variables
-12y^2+y=0
a = -12; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-12)·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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-12}=\frac{-2}{-24} =1/12 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-12}=\frac{0}{-24} =0 $
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