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(5y+2)-(16/3y+1)=8
We move all terms to the left:
(5y+2)-(16/3y+1)-(8)=0
Domain of the equation: 3y+1)!=0We get rid of parentheses
y∈R
5y-16/3y+2-1-8=0
We multiply all the terms by the denominator
5y*3y+2*3y-1*3y-8*3y-16=0
Wy multiply elements
15y^2+6y-3y-24y-16=0
We add all the numbers together, and all the variables
15y^2-21y-16=0
a = 15; b = -21; c = -16;
Δ = b2-4ac
Δ = -212-4·15·(-16)
Δ = 1401
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-\sqrt{1401}}{2*15}=\frac{21-\sqrt{1401}}{30} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{1401}}{2*15}=\frac{21+\sqrt{1401}}{30} $
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