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1/3x+2=5/2x
We move all terms to the left:
1/3x+2-(5/2x)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 2x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
1/3x-(+5/2x)+2=0
We get rid of parentheses
1/3x-5/2x+2=0
We calculate fractions
2x/6x^2+(-15x)/6x^2+2=0
We multiply all the terms by the denominator
2x+(-15x)+2*6x^2=0
Wy multiply elements
12x^2+2x+(-15x)=0
We get rid of parentheses
12x^2+2x-15x=0
We add all the numbers together, and all the variables
12x^2-13x=0
a = 12; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·12·0
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*12}=\frac{0}{24} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*12}=\frac{26}{24} =1+1/12 $
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