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3/4y=1/3y+4
We move all terms to the left:
3/4y-(1/3y+4)=0
Domain of the equation: 4y!=0
y!=0/4
y!=0
y∈R
Domain of the equation: 3y+4)!=0We get rid of parentheses
y∈R
3/4y-1/3y-4=0
We calculate fractions
9y/12y^2+(-4y)/12y^2-4=0
We multiply all the terms by the denominator
9y+(-4y)-4*12y^2=0
Wy multiply elements
-48y^2+9y+(-4y)=0
We get rid of parentheses
-48y^2+9y-4y=0
We add all the numbers together, and all the variables
-48y^2+5y=0
a = -48; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-48)·0
Δ = 25
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{25}=5$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-48}=\frac{-10}{-96} =5/48 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-48}=\frac{0}{-96} =0 $
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