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(13/3x)-(4/x)=12
We move all terms to the left:
(13/3x)-(4/x)-(12)=0
Domain of the equation: 3x)!=0
x!=0/1
x!=0
x∈R
Domain of the equation: x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
(+13/3x)-(+4/x)-12=0
We get rid of parentheses
13/3x-4/x-12=0
We calculate fractions
13x/3x^2+(-12x)/3x^2-12=0
We multiply all the terms by the denominator
13x+(-12x)-12*3x^2=0
Wy multiply elements
-36x^2+13x+(-12x)=0
We get rid of parentheses
-36x^2+13x-12x=0
We add all the numbers together, and all the variables
-36x^2+x=0
a = -36; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-36)·0
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-36}=\frac{-2}{-72} =1/36 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-36}=\frac{0}{-72} =0 $
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