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3/x+4=5/3x
We move all terms to the left:
3/x+4-(5/3x)=0
Domain of the equation: x!=0
x∈R
Domain of the equation: 3x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
3/x-(+5/3x)+4=0
We get rid of parentheses
3/x-5/3x+4=0
We calculate fractions
9x/3x^2+(-5x)/3x^2+4=0
We multiply all the terms by the denominator
9x+(-5x)+4*3x^2=0
Wy multiply elements
12x^2+9x+(-5x)=0
We get rid of parentheses
12x^2+9x-5x=0
We add all the numbers together, and all the variables
12x^2+4x=0
a = 12; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·12·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*12}=\frac{-8}{24} =-1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*12}=\frac{0}{24} =0 $
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