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12+2/3x=5+3/5x
We move all terms to the left:
12+2/3x-(5+3/5x)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 5x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
2/3x-(3/5x+5)+12=0
We get rid of parentheses
2/3x-3/5x-5+12=0
We calculate fractions
10x/15x^2+(-9x)/15x^2-5+12=0
We add all the numbers together, and all the variables
10x/15x^2+(-9x)/15x^2+7=0
We multiply all the terms by the denominator
10x+(-9x)+7*15x^2=0
Wy multiply elements
105x^2+10x+(-9x)=0
We get rid of parentheses
105x^2+10x-9x=0
We add all the numbers together, and all the variables
105x^2+x=0
a = 105; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·105·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*105}=\frac{-2}{210} =-1/105 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*105}=\frac{0}{210} =0 $
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