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3/2x+9=9/3x+6
We move all terms to the left:
3/2x+9-(9/3x+6)=0
Domain of the equation: 2x!=0
x!=0/2
x!=0
x∈R
Domain of the equation: 3x+6)!=0We get rid of parentheses
x∈R
3/2x-9/3x-6+9=0
We calculate fractions
9x/6x^2+(-18x)/6x^2-6+9=0
We add all the numbers together, and all the variables
9x/6x^2+(-18x)/6x^2+3=0
We multiply all the terms by the denominator
9x+(-18x)+3*6x^2=0
Wy multiply elements
18x^2+9x+(-18x)=0
We get rid of parentheses
18x^2+9x-18x=0
We add all the numbers together, and all the variables
18x^2-9x=0
a = 18; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·18·0
Δ = 81
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{81}=9$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*18}=\frac{0}{36} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*18}=\frac{18}{36} =1/2 $
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