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1/4y+10=1/7y
We move all terms to the left:
1/4y+10-(1/7y)=0
Domain of the equation: 4y!=0
y!=0/4
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
1/4y-(+1/7y)+10=0
We get rid of parentheses
1/4y-1/7y+10=0
We calculate fractions
7y/28y^2+(-4y)/28y^2+10=0
We multiply all the terms by the denominator
7y+(-4y)+10*28y^2=0
Wy multiply elements
280y^2+7y+(-4y)=0
We get rid of parentheses
280y^2+7y-4y=0
We add all the numbers together, and all the variables
280y^2+3y=0
a = 280; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·280·0
Δ = 9
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{9}=3$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*280}=\frac{-6}{560} =-3/280 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*280}=\frac{0}{560} =0 $
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