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5/6y+1=-1/2y+5
We move all terms to the left:
5/6y+1-(-1/2y+5)=0
Domain of the equation: 6y!=0
y!=0/6
y!=0
y∈R
Domain of the equation: 2y+5)!=0We get rid of parentheses
y∈R
5/6y+1/2y-5+1=0
We calculate fractions
10y/12y^2+6y/12y^2-5+1=0
We add all the numbers together, and all the variables
10y/12y^2+6y/12y^2-4=0
We multiply all the terms by the denominator
10y+6y-4*12y^2=0
We add all the numbers together, and all the variables
16y-4*12y^2=0
Wy multiply elements
-48y^2+16y=0
a = -48; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-48)·0
Δ = 256
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{256}=16$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-48}=\frac{-32}{-96} =1/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-48}=\frac{0}{-96} =0 $
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