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2/3y+y-4=32
We move all terms to the left:
2/3y+y-4-(32)=0
Domain of the equation: 3y!=0We add all the numbers together, and all the variables
y!=0/3
y!=0
y∈R
y+2/3y-36=0
We multiply all the terms by the denominator
y*3y-36*3y+2=0
Wy multiply elements
3y^2-108y+2=0
a = 3; b = -108; c = +2;
Δ = b2-4ac
Δ = -1082-4·3·2
Δ = 11640
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{11640}=\sqrt{4*2910}=\sqrt{4}*\sqrt{2910}=2\sqrt{2910}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-108)-2\sqrt{2910}}{2*3}=\frac{108-2\sqrt{2910}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-108)+2\sqrt{2910}}{2*3}=\frac{108+2\sqrt{2910}}{6} $
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