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1/5y-3=y
We move all terms to the left:
1/5y-3-(y)=0
Domain of the equation: 5y!=0We add all the numbers together, and all the variables
y!=0/5
y!=0
y∈R
-1y+1/5y-3=0
We multiply all the terms by the denominator
-1y*5y-3*5y+1=0
Wy multiply elements
-5y^2-15y+1=0
a = -5; b = -15; c = +1;
Δ = b2-4ac
Δ = -152-4·(-5)·1
Δ = 245
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{245}=\sqrt{49*5}=\sqrt{49}*\sqrt{5}=7\sqrt{5}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-7\sqrt{5}}{2*-5}=\frac{15-7\sqrt{5}}{-10} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+7\sqrt{5}}{2*-5}=\frac{15+7\sqrt{5}}{-10} $
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