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1/3y+3=1/7y.
We move all terms to the left:
1/3y+3-(1/7y.)=0
Domain of the equation: 3y!=0
y!=0/3
y!=0
y∈R
Domain of the equation: 7y.)!=0We add all the numbers together, and all the variables
y!=0/1
y!=0
y∈R
1/3y-(+1/7y.)+3=0
We get rid of parentheses
1/3y-1/7y.+3=0
We calculate fractions
7y/21y^2+(-3y)/21y^2+3=0
We multiply all the terms by the denominator
7y+(-3y)+3*21y^2=0
Wy multiply elements
63y^2+7y+(-3y)=0
We get rid of parentheses
63y^2+7y-3y=0
We add all the numbers together, and all the variables
63y^2+4y=0
a = 63; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·63·0
Δ = 16
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{16}=4$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*63}=\frac{-8}{126} =-4/63 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*63}=\frac{0}{126} =0 $
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